高手快来C++编写一个程序，用户输入一年（year）、月（month）、日（day），计算该日期对应的是该年的第几_百度知道
高手快来C++编写一个程序，用户输入一年（year）、月（month）、日（day），计算该日期对应的是该年的第几
VC2008版本，
题目是：用户输入一年（year）、月（month）、日（day），计算该日期对应的是该年的第几天。
我有更好的答案
我刚写的#include &iostream&int month_days[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
//每月的天数bool isLeap(int year)
//判断是否闰年{ if( (year%4 == 0 && year%100 != 0) || (year%400 == 0) )}int main(){ int year, month, while(true) {
printf(&请输入年月日：\n&);
scanf(&%d%d%d&, &year, &month, &day);
int days = 0;
for(int i = 1; i & i++)
//计算从1月到month-1月的总天数
days += month_days[i];
if(i == 2 && isLeap(year) == true)
days += 1;
//加上当月从1号到该天的天数
printf(&该日期对应的是该年的第 %d 天\n\n&,days); } return 0;}输入格式如：
采纳率：50%
/*思路：我们可以划分成三个函数来写：输入参数的检测函数、计算闰年的函数和计算日期的函数；最后写个主程序就OK了*///定义判断年份、月份和日期数据输入是否正确的函数bool judgemonthday(int m,int d,bool r){//判断数据输入，即月份和日期是否真确，rmday为真是真确，假为不正确 if(m&1 || m&12 || d&1 || d&31) {cout&&&数据输入错误。月份必须在1～12之间，日必须在1～31之间。&&&rmday=} else if(r==false && m==2 && d&28)
{cout&&&该年不是闰年，因而2月份没有29号&&&rmday=}
else if(r==true && m==2 && d&29)
{cout&&&该年是闰年，但2月份没有30、31号&&&rmday=} else if((m==4 || m==6 || m==9 || m==11) && d==31)
{cout&&m&&&月份没有31号&&&rmday=} else rmday=}//定义判断是否为闰年的函数bool judgeyear(int y){//r为真，即为闰年，r为假不为闰年 if(y%4==0 && y%100!=0) r= else if(y%400==0)
r=}//定义计算当前日期是该年第几天的函数int nday(int month,int day,bool r){ int i,s=0; for (i=1;i&i++)
switch(i) {
case 1: case 3: case 5: case 7: case 8: case 10:case 12:s+=31;
case 4: case 6: case 9: case 11: s+=30;
case 2:{if (r) s+=29;else s+=28;} } return s+}//主程序# include &iostream.h&void main(){ bool judgeyear(int y);//声明判断是否为闰年的函数 bool judgemonthday(int m,int d,bool r);//声明判断数据输入是否正确的函数 int
nday(int m,int d,bool r);//声明计算日期的函数 int year,month,day,n; bool ryear,//ryear是用来判断该年是否为闰年，rmd是用来判断输入的月份值、日值是否正确。 cout&&&请输入日期的年、月、日的值：&; cin&&year&&month&& ryear=judgeyear(year); rmd=judgemonthday(month,day,ryear);
if(rmd)//只有数据输入正确的情况下，才进行计算 {
n=nday(month,day,ryear);
cout&&year&&&年&&&month&&&月&&&day&&&日是&&&year&&&年的第&&&n&&&天。&&& }}程序试运行结果：请输入日期的年、月、日的值：日是2000年的第61天。
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#include&stdio.h&main(){ int day,month,year,sum, printf(&please input year,month,day:\n&); scanf(&%d,%d,%d&,&year,&month,&day); switch(month)/*先计算某月以前月份的总天数*/ { case 1:sum=0; case 2:sum=31; case 3:sum=59; case 4:sum=90; case 5:sum=120; case 6:sum=151; case 7:sum=181; case 8:sum=212; case 9:sum=243; case 10:sum=273; case 11:sum=304; case 12:sum=334; default:printf(&data error!\n&); } sum=sum+ /*再加上某天的天数*/ if(year%400==0||(year%4==0&&year%100!=0))/*判断是不是闰年*/
leap=1; else
leap=0; if(leap==1&&month&2)/*如果是闰年且月份大于2,总天数应该加一天*/
sum++;printf(&It is the %d day.\n&,sum);}
#include &iostream.h&#include &string.h&int main(){char a[255] = , b[255] = ;cout && &input first string: &;cin &&cout && &input second string: &;cin &&if(strlen(a) == strlen(b)){for(i = 0; i & (int)strlen(a); i++)if(a[i] != b[i])}if(i == (int)strlen(a))cout && &a equal to b& &&elsecout && &a not equal to b& &&return 0;}
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我们会通过消息、邮箱等方式尽快将举报结果通知您。C语言，请问这个错在哪里？_百度知道
C语言，请问这个错在哪里？
#include&stdio.h&
#include&stdlib.h&
int main()
int day,month,year,sum,
printf(&\nplease input year,month,day\n&);
scanf(&%d,%d,%d&,&year,&month,&day);
switch(month)
　case 1:sum=0;
　case 2:sum=31;
　case 3:...
我有更好的答案
请复制我的代码，你的代码不小心输入了汉字的空格导致一堆编译错误：#include&stdio.h&#include&stdlib.h&int&main(){&&&&int&day,month,year,sum,&&&&printf(&\nplease&input&year,month,day\n&);&&&&scanf(&%d,%d,%d&,&year,&month,&day);&&&&switch(month)&&&&{&&&&case&1:sum=0;&&&&case&2:sum=31;&&&&case&3:sum=59;&&&&case&4:sum=90;&&&&case&5:sum=120;&&&&case&6:sum=151;&&&&case&7:sum=181;&&&&case&8:sum=212;&&&&case&9:sum=243;&&&&case&10:sum=273;&&&&case&11:sum=304;&&&&case&12:sum=334;&&&&default:&&&&&&&&printf(&data&error&);&&&&&&&&&&&&}&&&&sum=sum+&&&&if((year%400==0)||(year%4==0&&year%100!=0))&&&&&&&&leap=1;&&&&else&&&&&&&&leap=0;&&&&if(leap==1&&month&2)&&&&&&&&sum++;&&&&printf(&It&is&the&%d&th&day.&,sum);&&&&system(&PAUSE&);&&&&return&0;}
采纳率：63%
来自团队：
你的代码语法上没有错误，请把输入法关掉，然后用英语重新打一次。就可以了。如果不想打，也可以复制我一下的代码#include&stdio.h&#include&stdlib.h&int main(){ int day,month,year,sum, printf(&请输入year,month,day\n&); scanf(&%d,%d,%d&,&year,&month,&day); switch(month) { case 1:sum=0; case 2:sum=31; case 3:sum=59; case 4:sum=90; case 5:sum=120; case 6:sum=151; case 7:sum=181; case 8:sum=212; case 9:sum=243; case 10:sum=273; case 11:sum=304; case 12:sum=334; default:printf(&data error\n&); } if((year%400==0)||(year%4==0&&year%100!=0))
leap=1; else
leap=0; if(leap==1&&month&12)
sum++; printf(&it is the %d th day&,sum); system(&PAUSE&); return 0;}
这种错误还跟输入法有关啊？以后该怎么避免呢？谢谢。
以后输入的时候看看着桌面（一般在右下角），看看现在是什么输入状态，是中文的话就按ctrl+shift调到英文状态。在输入的是后，你可一留意一下符号，英文和中文的符号是不一样的。要是都不确定的话，这一条方法一定行，打开控制面板——&语言，删除所有中文输入法。打完代码在添加回来。
Thank you.
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Leap Year 观后感
来源：互联网
贡献&责任编辑：鲁倩 &
True Love Comes Only for One Time----Impression of the Film Leap Year
I have watched the movie Leap Year today.
This is an ordinary love story like other movies I have watched. But I still got something special from the movie. What I have got is that love comes out by chance. film tells about several stories between the heroines Anna Brady and the hero Declan. A sweet beautiful young woman Anna wants to marry to a perfect man as her husband. One day, she had a date with her four-year boyfriend Jeremy, who worked as a high-powered heart surgeon. She was pleasant, attentive, and presentable and shared her goal of buying a condo in the best building in Boston. She had hoped that her boyfriend would ask her to marry him. But Jeremy presented her with a gift of twin diamond stud earrings instead of something she can wear on her finger. Anna was very disappointed. After dashing her hopes yet again, Jeremy hurried off to Dublin for a doctor’s conference.
Anna’s father knew that. He told his daughter that there was an old myth that decrees on Leap Day, and a woman in Ireland can propose to a man. So she booked a ticket and left for Dublin by air. Unfortunately, bad weather intervened and the plane is diverted to Cardiff. Despite canceled ferry boats, she makes her way to Ireland by hiring a tugboat. The skipper says they can't land at Cork but must head for Dingle, and Dingle in Ireland is more or less as far as she can get from Dublin. Then she came upon Declan.
Declan was the owner of a local pub and the taxi driver. Anna is now wet and tired but still plucky. In the pub, she asks Declan how she can get to Dublin. Declan offers to come to her rescue for the right price. Then they began their trip.
They argued all the way to Dublin, and Anna seemed like to be in dad luck all the time. Their car broke down and they missed the train to Dublin. They lived in a country house with two old couples, and they made their first kiss and shared the same bedroom. During the trip, the two begin to fall under each other's spell. Finally, they got to the bus station, and got on the bus to Dublin together. On February 29th afternoon, they arrived in Dublin and met Anna’s boyfriend Jeremy.
Jeremy asked Anna to marry him and Anna answered yes. Declan heard the answer and went back to his home. When Anna and Jeremy married, Anna was not happy. She realized that she fall in love with Declan. So she went to the small town and asked Declan to marry her. Finally, they lived together.
It didn’t mention whether the two were marry, but I guess they were very happy. Anna and Jeremy are supposed to get married at first, but Jeremy has no more care about this. I don’t think Jeremy truly love Anna, for he doesn’t worry about marrying Anna, and he doesn’t care about what Anna think, or he doesn’t care anybody’s thoughts. He is a self-centered man. At their wedding, when the alarm rang, he was busy gathering up his own things but not Anna.
Declan is different. He is a handsome calm considerate man. Although he often argued with Anna, he thought about Anna all the time. He followed Anna even had no cars, and told Anna the story of the old castle. He made Anna sleep on the only bed while himself in the bath, and they cooked dinner together. He walked with Anna in the rain, and gave his coat to Anna to keep her warm. When he heard Anna agreed to marry to Jeremy, he left without any word.
Comparing the two guys and remembering what had happened between them, Anna realized that the guy she truly loved was Declan. She found her heart coning back to that small town. Finally, she made her own right choose.
When love comes, one cannot realize it immediately. But he would know it sooner or later. And when love comes, don’t refuse it, because true love comes only for one time.以下内容为系统自动转化的文字版，可能排版等有问题，仅供您参考：True Love Comes Only for One Time ----Impression of the Film Leap YearI have watched the movie Leap Year today. This is an ordinary love storylike other movies I have watched. But I still got something special from the movie. What I have got is that love comes out by chance. The film tells about several stories between the heroines Anna Brady and the hero Declan. A sweet beautiful young woman Anna wants to marry to a perfect man as her husband. One day, she had a date with her four-year boyfriend Jeremy, who worked as a high-powered heart surgeon. She was pleasant, attentive, and presentable and shared her goal of buying a condo in the best building in Boston. She had hoped that her boyfriend would ask her to marry him. But Jeremy presented her with a gift of twin diamond stud earrings instead of something she can wear on her finger. Anna was very disappointed. After dashing her hopes yet again, Jeremy hurried off to Dublin for a doctor’s conference. Anna’s father knew that. He told his daughter that there was an old myth that decrees on Leap Day, and a woman in Ireland can propose to a man. So she booked a ticket and left for Dublin by air. Unfortunately, bad weather intervened and the plane is diverted to Cardiff. Despite canceled ferry boats, she makes her way to Ireland by hiring a tugboat. The skipper says they can't land at Cork but must head for Dingle, and Dingle in Ireland is more or less as far as she can get from Dublin. Then she came upon Declan. Declan was the owner of a local pub and the taxi driver. Anna is now wet and tired but still plucky. In the pub, she asks Declan how she can get to Dublin. Declan offers to come to her rescue for the right price. Then they began their trip. They argued all the way to Dublin, and Anna seemed like to be in dad luck all the time. Their car broke down and they missed the train to Dublin. They lived in a country house with two old couples, and they made their first kissand shared the same bedroom. During the trip, the two begin to fall under each other's spell. Finally, they got to the bus station, and got on the bus to Dublin together. On February 29th afternoon, they arrived in Dublin and met Anna’s boyfriend Jeremy. Jeremy asked Anna to marry him and Anna answered yes. Declan heard the answer and went back to his home. When Anna and Jeremy married, Anna was not happy. She realized that she fall in love with Declan. So she went to the small town and asked Declan to marry her. Finally, they lived together. It didn’t mention whether the two were marry, but I guess they were very happy. Anna and Jeremy are supposed to get married at first, but Jeremy has no more care about this. I don’t think Jeremy truly love Anna, for he doesn’t worry about marrying Anna, and he doesn’t care about what Anna think, or he doesn’t care anybody’s thoughts. He is a self-centered man. At their wedding, when the alarm rang, he was busy gathering up his own things but not Anna. Declan is different. He is a handsome calm considerate man. Although he often argued with Anna, he thought about Anna all the time. He followed Anna even had no cars, and told Anna the story of the old castle. He made Anna sleep on the only bed while himself in the bath, and they cooked dinner together. He walked with Anna in the rain, and gave his coat to Anna to keep her warm. When he heard Anna agreed to marry to Jeremy, he left without any word. Comparing the two guys and remembering what had happened between them, Anna realized that the guy she truly loved was Declan. She found her heart coning back to that small town. Finally, she made her own right choose. When love comes, one cannot realize it immediately. But he would know it sooner or later. And when love comes, don’t refuse it, because true love comes only for one time.
与《》相关：
- Copyright & 2017 www.xue63.com All Rights Reservedint day_of_year(int year,int month,int day) 哪位高手指点哪错了？谢~~_百度知道
int day_of_year(int year,int month,int day) 哪位高手指点哪错了？谢~~
int day_of_year(int year,int month,int day)
int tab[2][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31}
{0,31,29,31,30,31,30,31,31,30,31,30,31}
leap=(year%4==0&&year%100==0) || year%400==0;
for(k=0;k&k++...
我有更好的答案
leap=(year%4==0&&year%100==0) || year%400==0;这句闰年的判断语句错了.改成leap=(year%4==0&&year%100!=0) || year%400==0;或者leap=year%4==0&&(year%100!=0 || year%400==0);因为闰年的判断条件是能被4整除,逢百年就是能被400整除或者不能被100整除.
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我们会通过消息、邮箱等方式尽快将举报结果通知您。C语言程序设计50例(经典收藏)
转载 &更新时间：日 09:13:11 & 作者：
本篇文章是对C语言程序设计的50个小案例进行了详细的分析介绍，需要的朋友参考下
【程序1】题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去　　　　　　掉不满足条件的排列。 2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j,k;& printf("\n");& for(i=1;i&5;i++) /*以下为三重循环*/&&& for(j=1;j&5;j++)&&&&& for (k=1;k&5;k++)&&&&& {&&&&&&& if (i!=k&&i!=j&&j!=k) /*确保i、j、k三位互不相同*/&&&&&&& printf("%d,%d,%d\n",i,j,k);&&&&& }& getch();}==============================================================【程序2】题目：企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10%；利润高　　　于10万元，低于20万元时，低于10万元的部分按10%提成，高于10万元的部分，可可提　　　成7.5%；20万到40万之间时，高于20万元的部分，可提成5%；40万到60万之间时高于　　　40万元的部分，可提成3%；60万到100万之间时，高于60万元的部分，可提成1.5%，高于　　　100万元时，超过100万元的部分按1%提成，从键盘输入当月利润I，求应发放奖金总数？1.程序分析：请利用数轴来分界，定位。注意定义时需把奖金定义成长整型。　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){&& int bonus1,bonus2,bonus4,bonus6,bonus10,& scanf("%ld",&i); & bonus1=. 1;& bonus2=bonus1+.75;& bonus4=bonus2+.5;& bonus6=bonus4+.3;& bonus10=bonus6+.15;& if(i&=100000)&&& bonus=i*0.1;&&& else if(i&=200000)&&&&& bonus=bonus1+(i-.075;&&&&&&& else if(i&=400000)&&&&&&&&& bonus=bonus2+(i-.05;&&&&&&&&&&& else if(i&=600000)&&&&&&&&&&&&& bonus=bonus4+(i-.03;&&&&&&&&&&&&&&& else if(i&=1000000)&&&&&&&&&&&&&&&&& bonus=bonus6+(i-.015;&&&&&&&&&&&&&&&&&&& else&&&&&&&&&&&&&&&&&&&&& bonus=bonus10+(i-.01;& printf("bonus=%d",bonus);& getch(); }==============================================================【程序3】题目：一个整数，它加上100后是一个完全平方数，再加上168又是一个完全平方数，请问该数是多少？1.程序分析：在10万以内判断，先将该数加上100后再开方，再将该数加上268后再开方，如果开方后　　　　　　的结果满足如下条件，即是结果。请看具体分析：2.程序源代码： 代码如下:#include "math.h"#include "stdio.h"#include "conio.h"main(){& long int i,x,y,z;& for (i=1;i&100000;i++)& {&&& x=sqrt(i+100); /*x为加上100后开方后的结果*/&&& y=sqrt(i+268); /*y为再加上168后开方后的结果*/&&& if(x*x==i+100&&y*y==i+268) /*如果一个数的平方根的平方等于该数，这说明此数是完全平方数*/&&& printf("\n%ld\n",i);& }& getch();}==============================================================【程序4】题目：输入某年某月某日，判断这一天是这一年的第几天？1.程序分析：以3月5日为例，应该先把前两个月的加起来，然后再加上5天即本年的第几天，特殊　　　　　　情况，闰年且输入月份大于3时需考虑多加一天。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int day,month,year,sum,& printf("\nplease input year,month,day\n");& scanf("%d,%d,%d",&year,&month,&day);& switch(month) /*先计算某月以前月份的总天数*/& {&&& case 1:sum=0;&&& case 2:sum=31;&&& case 3:sum=59;&&& case 4:sum=90;&&& case 5:sum=120;&&& case 6:sum=151;&&& case 7:sum=181;&&& case 8:sum=212;&&& case 9:sum=243;&&& case 10:sum=273;&&& case 11:sum=304;&&& case 12:sum=334;&&& default:printf("data error");& }& sum=sum+ /*再加上某天的天数*/& if(year%400==0||(year%4==0&&year%100!=0)) /*判断是不是闰年*/&&& leap=1;& else&&& leap=0;& if(leap==1&&month&2) /*如果是闰年且月份大于2,总天数应该加一天*/&&& sum++;& printf("It is the %dth day.",sum);& getch(); }==============================================================【程序5】题目：输入三个整数x,y,z，请把这三个数由小到大输出。1.程序分析：我们想办法把最小的数放到x上，先将x与y进行比较，如果x&y则将x与y的值进行交换，　　　　　　然后再用x与z进行比较，如果x&z则将x与z的值进行交换，这样能使x最小。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int x,y,z,t;& scanf("%d%d%d",&x,&y,&z);& if (x&y)&&& {t=x;x=y;y=t;} /*交换x,y的值*/& if(x&z)&&& {t=z;z=x;x=t;} /*交换x,z的值*/& if(y&z)&&& {t=y;y=z;z=t;} /*交换z,y的值*/& printf("small to big: %d %d %d\n",x,y,z);& getch(); }==============================================================【程序6】题目：用*号输出字母C的图案。1.程序分析：可先用'*'号在纸上写出字母C，再分行输出。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& printf("Hello C-world!\n");& printf(" ****\n");& printf(" *\n");& printf(" * \n");& printf(" ****\n");& getch(); }==============================================================【程序7】题目：输出特殊图案，请在c环境中运行，看一看，Very Beautiful!1.程序分析：字符共有256个。不同字符，图形不一样。　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& char a=176,b=219;& printf("%c%c%c%c%c\n",b,a,a,a,b);& printf("%c%c%c%c%c\n",a,b,a,b,a);& printf("%c%c%c%c%c\n",a,a,b,a,a);& printf("%c%c%c%c%c\n",a,b,a,b,a);& printf("%c%c%c%c%c\n",b,a,a,a,b);& getch(); }==============================================================【程序8】题目：输出9*9口诀。1.程序分析：分行与列考虑，共9行9列，i控制行，j控制列。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j,& printf("\n");& for (i=1;i&10;i++)& {&&& for(j=1;j&10;j++)&&& {&&&&& result=i*j;&&&&& printf("%d*%d=%-3d",i,j,result); /*-3d表示左对齐，占3位*/&&& }&&& printf("\n"); /*每一行后换行*/& }& getch();}==============================================================【程序9】题目：要求输出国际象棋棋盘。1.程序分析：用i控制行，j来控制列，根据i+j的和的变化来控制输出黑方格，还是白方格。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j;& for(i=0;i&8;i++)& {&&& for(j=0;j&8;j++)&&&&& if((i+j)%2==0)&&&&&&& printf("%c%c",219,219);&&&&& else&&&&&&& printf("& ");&&& printf("\n");& }& getch(); }============================================================== 【程序10】题目：打印楼梯，同时在楼梯上方打印两个笑脸。 1.程序分析：用i控制行，j来控制列，j根据i的变化来控制输出黑方格的个数。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j;& printf("\1\1\n"); /*输出两个笑脸*/& for(i=1;i&11;i++)& {&&& for(j=1;j&=i;j++)&&&&& printf("%c%c",219,219);&&& printf("\n");& }& getch(); }【程序11】题目：古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月　　　后每个月又生一对兔子，假如兔子都不死，问每个月的兔子总数为多少？1.程序分析：　兔子的规律为数列1,1,2,3,5,8,13,21....2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& long f1,f2;&& f1=f2=1;& for(i=1;i&=20;i++)& {&&& printf("%12ld %12ld",f1,f2);&&& if(i%2==0) printf("\n"); /*控制输出，每行四个*/&&& f1=f1+f2; /*前两个月加起来赋值给第三个月*/&&& f2=f1+f2; /*前两个月加起来赋值给第三个月*/& }& getch();}==============================================================【程序12】题目：判断101-200之间有多少个素数，并输出所有素数。1.程序分析：判断素数的方法：用一个数分别去除2到sqrt(这个数)，如果能被整除，　　　　　　则表明此数不是素数，反之是素数。 　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"#include "math.h"main(){& int m,i,k,h=0,leap=1;& printf("\n");& for(m=101;m&=200;m++)& {&&& k=sqrt(m+1);&&& for(i=2;i&=k;i++)&&& if(m%i==0)&&& {&&&&& leap=0;&&&&&&&& }&&& if(leap)&&& {&&&&& printf("%-4d",m);&&&&& h++;&&&&& if(h%10==0)&&&&&&& printf("\n");&&& }&&& leap=1;& }& printf("\nThe total is %d",h);& getch();}==============================================================【程序13】题目：打印出所有的“水仙花数”，所谓“水仙花数”是指一个三位数，其各位数字立方和等于该数　　　本身。例如：153是一个“水仙花数”，因为153=1的三次方＋5的三次方＋3的三次方。1.程序分析：利用for循环控制100-999个数，每个数分解出个位，十位，百位。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j,k,n;& printf("'water flower'number is:");& for(n=100;n&1000;n++)& {&&& i=n/100;/*分解出百位*/&&& j=n/10%10;/*分解出十位*/&&& k=n%10;/*分解出个位*/&&& if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)&&&&& printf("%-5d",n);& }& getch();}==============================================================【程序14】题目：将一个正整数分解质因数。例如：输入90,打印出90=2*3*3*5。程序分析：对n进行分解质因数，应先找到一个最小的质数k，然后按下述步骤完成：(1)如果这个质数恰等于n，则说明分解质因数的过程已经结束，打印出即可。(2)如果n&&k，但n能被k整除，则应打印出k的值，并用n除以k的商,作为新的正整数你n,　重复执行第一步。(3)如果n不能被k整除，则用k+1作为k的值,重复执行第一步。2.程序源代码： 代码如下:/* zheng int is divided yinshu*/#include "stdio.h"#include "conio.h"main(){& int n,i;& printf("\nplease input a number:\n");& scanf("%d",&n);& printf("%d=",n);& for(i=2;i&=n;i++)&&& while(n!=i)&&& {&&&&& if(n%i==0)&&&&& {&&&&&&& printf("%d*",i);&&&&&&& n=n/i;&&&&& }&&&&& else&&&&&&&&&& }& printf("%d",n);& getch();}==============================================================【程序15】题目：利用条件运算符的嵌套来完成此题：学习成绩&=90分的同学用A表示，60-89分之间的用B表示，　　　60分以下的用C表示。1.程序分析：(a&b)?a:b这是条件运算符的基本例子。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){&&& printf("please input a score\n");& scanf("%d",&score);& grade=score&=90?'A':(score&=60?'B':'C');& printf("%d belongs to %c",score,grade);& getch();}==============================================================【程序16】题目：输入两个正整数m和n，求其最大公约数和最小公倍数。1.程序分析：利用辗除法。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int a,b,num1,num2,& printf("please input two numbers:\n");& scanf("%d,%d",&num1,&num2);& if(num1&num2)/*交换两个数，使大数放在num1上*/& {&&& temp=num1;&&& num1=num2;&&& num2=& }& a=num1;b=num2;& while(b!=0)/*利用辗除法，直到b为0为止*/& {&&& temp=a%b;&&& a=b;&&& b=& }& printf("gongyueshu:%d\n",a);& printf("gongbeishu:%d\n",num1*num2/a);& getch();}==============================================================【程序17】题目：输入一行字符，分别统计出其中英文字母、空格、数字和其它字符的个数。1.程序分析：利用while语句,条件为输入的字符不为'\n'.　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){&& int letters=0,space=0,digit=0,others=0;& printf("please input some characters\n");& while((c=getchar())!='\n')& {&&& if(c&='a'&&c&='z'||c&='A'&&c&='Z')&&&&& letters++;&&&&& else if(c==' ')&&&&&&& space++;&&&&&&& else if(c&='0'&&c&='9')&&&&&&&&& digit++;&&&&&&& else&&&&&&&&& others++;& }& printf("all in all:char=%d space=%d digit=%d others=%d\n",letters,& space,digit,others);& getch();}==============================================================【程序18】题目：求s=a+aa+aaa+aaaa+aa...a的值，其中a是一个数字。例如2+22+222+(此时　　　共有5个数相加)，几个数相加有键盘控制。1.程序分析：关键是计算出每一项的值。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int a,n,count=1;& long int sn=0,tn=0;& printf("please input a and n\n");& scanf("%d,%d",&a,&n);& printf("a=%d,n=%d\n",a,n);& while(count&=n)& {&&& tn=tn+a;&&& sn=sn+&&& a=a*10;&&& ++& }& printf("a+aa+...=%ld\n",sn);& getch();}==============================================================【程序19】题目：一个数如果恰好等于它的因子之和，这个数就称为“完数”。例如6=1＋2＋3.编程　　　找出1000以内的所有完数。1. 程序分析：请参照程序&--上页程序14.2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& static int k[10];& int i,j,n,s;& for(j=2;j&1000;j++)& {&&& n=-1;&&& s=j;&&& for(i=1;i&j;i++)&&& {&&&&& if((j%i)==0)&&&&& {&&&&&&& n++;&&&&&&& s=s-i;&&&&&&& k[n]=i;&&&&& }&&& }&&& if(s==0)&&& {&&&&& printf("%d is a wanshu",j);&&&&& for(i=0;i&n;i++)&&&&& printf("%d,",k);&&&&& printf("%d\n",k[n]);&&& }& }& getch();}==============================================================【程序20】题目：一球从100米高度自由落下，每次落地后反跳回原高度的一半；再落下，求它在　　　第10次落地时，共经过多少米？第10次反弹多高？1.程序分析：见下面注释2.程序源代码： 代码如下:#include "stdio.h"#include "stdio.h"main(){& float sn=100.0,hn=sn/2;&& for(n=2;n&=10;n++)& {&&& sn=sn+2*/*第n次落地时共经过的米数*/&&& hn=hn/2; /*第n次反跳高度*/& }& printf("the total of road is %f\n",sn);& printf("the tenth is %f meter\n",hn);& getch();}【程序21】题目：猴子吃桃问题：猴子第一天摘下若干个桃子，当即吃了一半，还不瘾，又多吃了一个　　　第二天早上又将剩下的桃子吃掉一半，又多吃了一个。以后每天早上都吃了前一天剩下　　　的一半零一个。到第10天早上想再吃时，见只剩下一个桃子了。求第一天共摘了多少。1.程序分析：采取逆向思维的方法，从后往前推断。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int day,x1,x2;& day=9;& x2=1;& while(day&0)& {&&& x1=(x2+1)*2;/*第一天的桃子数是第2天桃子数加1后的2倍*/&&& x2=x1;&&& day--;& }& printf("the total is %d\n",x1);& getch();}==============================================================【程序22】题目：两个乒乓球队进行比赛，各出三人。甲队为a,b,c三人，乙队为x,y,z三人。已抽签决定　　　比赛名单。有人向队员打听比赛的名单。a说他不和x比，c说他不和x,z比，请编程序找出　　　三队赛手的名单。1.程序分析：判断素数的方法：用一个数分别去除2到sqrt(这个数)，如果能被整除，　　　　　　则表明此数不是素数，反之是素数。 　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& char i,j,k;/*i是a的对手，j是b的对手，k是c的对手*/& for(i='x';i&='z';i++)&&& for(j='x';j&='z';j++)&&& {&&&&& if(i!=j)&&&&& for(k='x';k&='z';k++)&&&&& {&&&&&&& if(i!=k&&j!=k)&&&&&&& {&&&&&&&&& if(i!='x'&&k!='x'&&k!='z')&&&&&&&&&&& printf("order is a--%c\tb--%c\tc--%c\n",i,j,k);&&&&&&&& }&&&&& }&&& }& getch();}==============================================================【程序23】题目：打印出如下图案（菱形）&& *& ********************& ***&& *1.程序分析：先把图形分成两部分来看待，前四行一个规律，后三行一个规律，利用双重　　　　　　for循环，第一层控制行，第二层控制列。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,j,k;& for(i=0;i&=3;i++)& {&&& for(j=0;j&=2-i;j++)&&&&& printf(" ");&&& for(k=0;k&=2*i;k++)&&&&& printf("*");&&& printf("\n");& }& for(i=0;i&=2;i++)& {&&& for(j=0;j&=i;j++)&&&&& printf(" ");&&& for(k=0;k&=4-2*i;k++)&&&&& printf("*");&&& printf("\n");& }& getch();}==============================================================【程序24】题目：有一分数序列：2/1，3/2，5/3，8/5，13/8，21/13...求出这个数列的前20项之和。1.程序分析：请抓住分子与分母的变化规律。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int n,t,number=20;& float a=2,b=1,s=0;& for(n=1;n&=n++)& {&&& s=s+a/b;&&& t=a;a=a+b;b=t;/*这部分是程序的关键，请读者猜猜t的作用*/& }& printf("sum is %9.6f\n",s);& getch();}==============================================================【程序25】题目：求1+2!+3!+...+20!的和1.程序分析：此程序只是把累加变成了累乘。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& float n,s=0,t=1;& for(n=1;n&=20;n++)& {&&& t*=n;&&& s+=t;& }& printf("1+2!+3!...+20!=%e\n",s);& getch();}==============================================================【程序26】题目：利用递归方法求5!。1.程序分析：递归公式：fn=fn_1*4!2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){&& int fact();& for(i=0;i&5;i++)& printf("\40:%d!=%d\n",i,fact(i));& getch();}int fact(j){&& if(j==0)&&& sum=1;& else&&& sum=j*fact(j-1);&}==============================================================【程序27】题目：利用递归函数调用方式，将所输入的5个字符，以相反顺序打印出来。1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i=5;& void palin(int n);& printf("\40:");& palin(i);& printf("\n");& getch();}void palin(n){&& if(n&=1)& {&&& next=getchar();&&& printf("\n\0:");&&& putchar(next);& }& else& {&&& next=getchar();&&& palin(n-1);&&& putchar(next);& }}==============================================================【程序28】题目：有5个人坐在一起，问第五个人多少岁？他说比第4个人大2岁。问第4个人岁数，他说比第　　　3个人大2岁。问第三个人，又说比第2人大两岁。问第2个人，说比第一个人大两岁。最后　　　问第一个人，他说是10岁。请问第五个人多大？1.程序分析：利用递归的方法，递归分为回推和递推两个阶段。要想知道第五个人岁数，需知道　　　　　　第四人的岁数，依次类推，推到第一人（10岁），再往回推。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"age(n){&& if(n==1) c=10;& else c=age(n-1)+2;& return(c);}main(){& printf("%d",age(5));& getch();}==============================================================【程序29】题目：给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出各位数字。1. 程序分析：学会分解出每一位数，如下解释：(这里是一种简单的算法，师专数002班赵鑫提供)2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main( ){& long a,b,c,d,e,x;& scanf("%ld",&x);& a=x/10000;/*分解出万位*/& b=x%;/*分解出千位*/& c=x%;/*分解出百位*/& d=x%100/10;/*分解出十位*/& e=x%10;/*分解出个位*/& if (a!=0) printf("there are 5, %ld %ld %ld %ld %ld\n",e,d,c,b,a);& else if (b!=0) printf("there are 4, %ld %ld %ld %ld\n",e,d,c,b);&&& else if (c!=0) printf(" there are 3,%ld %ld %ld\n",e,d,c);&&&&& else if (d!=0) printf("there are 2, %ld %ld\n",e,d);&&&&&&& else if (e!=0) printf(" there are 1,%ld\n",e);& getch();}==============================================================【程序30】题目：一个5位数，判断它是不是回文数。即12321是回文数，个位与万位相同，十位与千位相同。　　　1.程序分析：同29例2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main( ){& long ge,shi,qian,wan,x;& scanf("%ld",&x);& wan=x/10000;& qian=x%;& shi=x%100/10;& ge=x%10;& if(ge==wan&&shi==qian)/*个位等于万位并且十位等于千位*/&&& printf("this number is a huiwen\n");& else&&& printf("this number is not a huiwen\n");& getch();}【程序31】题目：请输入星期几的第一个字母来判断一下是星期几，如果第一个字母一样，则继续　　　判断第二个字母。1.程序分析：用情况语句比较好，如果第一个字母一样，则判断用情况语句或if语句判断第二个字母。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"void main(){&& printf("please input the first letter of someday\n");& while((letter=getch())!='Y')/*当所按字母为Y时才结束*/& {&&& switch (letter)&&& {&&&&& case 'S':printf("please input second letter\n");&&&&& if((letter=getch())=='a')&&&&&&& printf("saturday\n");&&&&&&& else if ((letter=getch())=='u')&&&&&&&&& printf("sunday\n");&&&&&&&&& else printf("data error\n");&&&&&&&&&& case 'F':printf("friday\n");&&&&& case 'M':printf("monday\n");&&&&& case 'T':printf("please input second letter\n");&&&&& if((letter=getch())=='u')&&&&&&& printf("tuesday\n");&&&&&&& else if ((letter=getch())=='h')&&&&&&&&& printf("thursday\n");&&&&&&& else printf("data error\n");&&&&&&&&&& case 'W':printf("wednesday\n");&&&&& default: printf("data error\n");&&& }& }& getch();}==============================================================【程序32】题目：Press any key to change color, do you want to try it. Please hurry up!1.程序分析：　　　　　　　　　　　　2.程序源代码： 代码如下:#include "conio.h"#include "stdio.h"void main(void){&& for (color = 0; color & 8; color++)& {&&& textbackground(color);/*设置文本的背景颜色*/&&& cprintf("This is color %d\r\n", color);&&& cprintf("Press any key to continue\r\n");&&& getch();/*输入字符看不见*/& }}==============================================================【程序33】题目：学习gotoxy()与clrscr()函数　　　1.程序分析：2.程序源代码： 代码如下:#include "conio.h"#include "stdio.h"void main(void){& clrscr();/*清屏函数*/& textbackground(2);& gotoxy(1, 5);/*定位函数*/& cprintf("Output at row 5 column 1\n");& textbackground(3);& gotoxy(20, 10);& cprintf("Output at row 10 column 20\n");& getch();}==============================================================【程序34】题目：练习函数调用1. 程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"void hello_world(void){& printf("Hello, world!\n");}void three_hellos(void){&& for (counter = 1; counter &= 3; counter++)&&& hello_world();/*调用此函数*/}void main(void){& three_hellos();/*调用此函数*/& getch();}==============================================================【程序35】题目：文本颜色设置1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"void main(void){&& for (color = 1; color & 16; color++)& {&&& textcolor(color);/*设置文本颜色*/&&& cprintf("This is color %d\r\n", color);& }& textcolor(128 + 15);& cprintf("This is blinking\r\n");& getch();}==============================================================【程序36】题目：求100之内的素数　　　1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "math.h"#define N 101main(){& int i,j,line,a[N];& for(i=2;i&N;i++) a=i;&&& for(i=2;i&sqrt(N);i++)&&&&& for(j=i+1;j&N;j++)&&&&& {&&&&&&& if(a!=0&&a[j]!=0)&&&&&&&&& if(a[j]%a==0)&&&&&&&&&&& a[j]=0;&&&&& }& printf("\n");& for(i=2,line=0;i&N;i++)& {&&& if(a!=0)&&& {&&&&& printf("%5d",a);&&&&& line++;&&& }&&& if(line==10)&&& {&&&&& printf("\n");&&&&& line=0;&&& }& }& getch();}==============================================================【程序37】题目：对10个数进行排序1.程序分析：可以利用选择法，即从后9个比较过程中，选择一个最小的与第一个元素交换，　　　　　　下次类推，即用第二个元素与后8个进行比较，并进行交换。 　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"#define N 10main(){& int i,j,min,tem,a[N];& /*input data*/& printf("please input ten num:\n");& for(i=0;i&N;i++)& {&&& printf("a[%d]=",i);&&& scanf("%d",&a);& }& printf("\n");& for(i=0;i&N;i++)&&& printf("%5d",a);& printf("\n");& /*sort ten num*/& for(i=0;i&N-1;i++)& {&&& min=i;&&& for(j=i+1;j&N;j++)&&&&& if(a[min]&a[j])&&&&&&& min=j;&&& tem=a;&&& a=a[min];&&& a[min]=& }& /*output data*/& printf("After sorted \n");& for(i=0;i&N;i++)& printf("%5d",a);& getch();}==============================================================【程序38】题目：求一个3*3矩阵对角线元素之和1.程序分析：利用双重for循环控制输入二维数组，再将a累加后输出。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"/* 如果使用的是TC系列编译器则可能需要添加下句 */static void dummyfloat(float *x){ dummyfloat(&y);}main(){& float a[3][3],sum=0;& int i,j;& printf("please input rectangle element:\n");& for(i=0;i&3;i++)&&& for(j=0;j&3;j++)&&&&& scanf("%f",&a[j]);& for(i=0;i&3;i++)&&& sum=sum+a;& printf("duijiaoxian he is %6.2f",sum);& getch();}==============================================================【程序39】题目：有一个已经排好序的数组。现输入一个数，要求按原来的规律将它插入数组中。1. 程序分析：首先判断此数是否大于最后一个数，然后再考虑插入中间的数的情况，插入后　　　　　此元素之后的数，依次后移一个位置。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int a[11]={1,4,6,9,13,16,19,28,40,100};& int temp1,temp2,number,end,i,j;& printf("original array is:\n");& for(i=0;i&10;i++)&&& printf("%5d",a);& printf("\n");& printf("insert a new number:");& scanf("%d",&number);& end=a[9];& if(number&end)&&& a[10]=& else& {&&& for(i=0;i&10;i++)&&& {&&&&& if(a&number)&&&&& {&&&&&&& temp1=a;&&&&&&& a=&&&&&&& for(j=i+1;j&11;j++)&&&&&&& {&&&&&&&&& temp2=a[j];&&&&&&&&& a[j]=temp1;&&&&&&&&& temp1=temp2;&&&&&&& }&&&&&&&&&&&& }&&& }& }& for(i=0;i&11;i++)&&& printf("%6d",a);& getch();}==============================================================【程序40】题目：将一个数组逆序输出。1.程序分析：用第一个与最后一个交换。2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"#define N 5main(){& int a[N]={9,6,5,4,1},i,& printf("\n original array:\n");& for(i=0;i&N;i++)&&& printf("%4d",a);& for(i=0;i&N/2;i++)& {&&& temp=a;&&& a=a[N-i-1];&&& a[N-i-1]=& }& printf("\n sorted array:\n");& for(i=0;i&N;i++)&&& printf("%4d",a);& getch();}==============================================================【程序41】题目：学习static定义静态变量的用法　　　1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"varfunc(){& int var=0;& static int static_var=0;& printf("\40:var equal %d \n",var);& printf("\40:static var equal %d \n",static_var);& printf("\n");& var++;& static_var++;}void main(){&& for(i=0;i&3;i++)&&& varfunc();& getch();}==============================================================【程序42】题目：学习使用auto定义变量的用法1.程序分析：　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,& num=2;& for(i=0;i&3;i++)& {&&& printf("\40: The num equal %d \n",num);&&& num++;&&& {&&&&& auto int num=1;&&&&& printf("\40: The internal block num equal %d \n",num);&&&&& num++;&&& }& }& getch();}==============================================================【程序43】题目：学习使用static的另一用法。　　　1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"main(){& int i,& num=2;& for(i=0;i&3;i++)& {&&& printf("\40: The num equal %d \n",num);&&& num++;&&& {&&&&& static int num=1;&&&&& printf("\40:The internal block num equal %d\n",num);&&&&& num++;&&& }& }& getch();}==============================================================【程序44】题目：学习使用external的用法。1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"int a,b,c;void add(){&& a=3;& c=a+b;}void main(){& a=b=4;& add();& printf("The value of c is equal to %d\n",c);& getch();}==============================================================【程序45】题目：学习使用register定义变量的方法。1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"void main(){&& int tmp=0;& for(i=1;i&=100;i++)& tmp+=i;& printf("The sum is %d\n",tmp);& getch();}==============================================================【程序46】题目：宏#define命令练习(1)　　　1.程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"#define TRUE 1#define FALSE 0#define SQ(x) (x)*(x)void main(){&& int again=1;& printf("\40: Program will stop if input value less than 50.\n");& while(again)& {&&& printf("\40:Please input number==&");&&& scanf("%d",&num);&&& printf("\40:The square for this number is %d \n",SQ(num));&&& if(num&=50)&&&&& again=TRUE;&&& else&&&&& again=FALSE;& }& getch();}==============================================================【程序47】题目：宏#define命令练习(2)1.程序分析：　　　　　　　　　　　　2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"/*宏定义中允许包含两道衣裳命令的情形，此时必须在最右边加上"\"*/#define exchange(a,b) { \&&&&&&&&&&&&&&&&&&&&&&&\&&&&&&&&&&&&&&&&&&&&&&& t=a;\&&&&&&&&&&&&&&&&&&&&&&& a=b;\&&&&&&&&&&&&&&&&&&&&&&& b=t;\&&&&&&&&&&&&&&&&&&&&& }void main(void){& int x=10;& int y=20;& printf("x=%d; y=%d\n",x,y);& exchange(x,y);& printf("x=%d; y=%d\n",x,y);& getch();}==============================================================【程序48】题目：宏#define命令练习(3)　　　1.程序分析：2.程序源代码： 代码如下:#define LAG &#define SMA &#define EQ ==#include "stdio.h"#include "conio.h"void main(){& int i=10;& int j=20;& if(i LAG j)&&& printf("\40: %d larger than %d \n",i,j);&&& else if(i EQ j)&&&&& printf("\40: %d equal to %d \n",i,j);&&&&& else if(i SMA j)&&&&&&& printf("\40:%d smaller than %d \n",i,j);&&&&& else&&&&&&& printf("\40: No such value.\n");& getch();}==============================================================【程序49】题目：#if #ifdef和#ifndef的综合应用。1. 程序分析：2.程序源代码： 代码如下:#include "stdio.h"#include "conio.h"#define MAX#define MAXIMUM(x,y) (x&y)?x:y#define MINIMUM(x,y) (x&y)?y:xvoid main(){& int a=10,b=20;#ifdef MAX& printf("\40: The larger one is %d\n",MAXIMUM(a,b));#else& printf("\40: The lower one is %d\n",MINIMUM(a,b));#endif#ifndef MIN& printf("\40: The lower one is %d\n",MINIMUM(a,b));#else& printf("\40: The larger one is %d\n",MAXIMUM(a,b));#endif#undef MAX#ifdef MAX& printf("\40: The larger one is %d\n",MAXIMUM(a,b));#else& printf("\40: The lower one is %d\n",MINIMUM(a,b));#endif#define MIN#ifndef MIN& printf("\40: The lower one is %d\n",MINIMUM(a,b));#else& printf("\40: The larger one is %d\n",MAXIMUM(a,b));#endif& getch();}
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