【求助】APP上文案显示没有空行是什么鬼_晋江文学城吧_百度贴吧
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【求助】APP上文案显示没有空行是什么鬼
以前都在电脑上看的从来没有注意过app，结果今天随便搜一下自己的文直接懵逼[○?‘Д??○]密密麻麻的文案我自己都看不下去，查了很久的攻略都没有找到代码中的问题，而且在电脑上分断什么的都是很正常的下图左边是显示出来的文案，右边是源代码[○?‘Д??○]
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app就这样儿，拼字么？
每段开头&p&结尾&/p& 试试 括号输入法打出来的不标准 自己在文案复制下
Σ(|||▽||| )其实我也是这样
用了三楼大大的办法有一丢丢改进……我决定上电脑把字体代码每一行都复制一遍了[○?‘Д??○]
不是在网页上修改一下 app就会好了吗
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使用签名档&&
保存至快速回贴每两个测试组之间有空行，最后一组数据后面没空行hdu2088
Box of Bricks
(Java/Others)&&&&Memory
K (Java/Others)
Total Submission(s):
6703&&&&Accepted
Submission(s): 2297
Problem Description
Little Bob likes playing with his box of bricks. He puts the
bricks one upon another and builds stacks of different height.
“Look, I've built a wall!”, he tells his older sister Alice. “Nah,
you should make all stacks the same height. Then you would have a
real wall.”, she retorts. After a little consideration, Bob sees
that she is right. So he sets out to rearrange the bricks, one by
one, such that all stacks are the same height afterwards. But since
Bob is lazy he wants to do this with the minimum number of bricks
moved. Can you help?
The input consists of several data sets. Each set begins with
a line containing the number n of stacks Bob has built. The next
line contains n numbers, the heights hi of the n stacks. You may
assume 1≤n≤50 and 1≤hi≤100.
The total number of bricks will be divisible by the number of
stacks. Thus, it is always possible to rearrange the bricks such
that all stacks have the same height.
The input is terminated by a set starting with n = 0. This set
should not be processed.
For each set, print the minimum number of bricks that have to
be moved in order to make all the stacks the same height.
Output a blank line between each set.
Sample Input
Sample Output
#include&cstdio&
#include&cstring&
#include&cstdlib&
#include&algorithm&
#include&cmath&
#include&iostream&
int a[1001];
int main(){
&int n,m,i,flag=0;
&memset(a,0,sizeof(a));
&while(cin&&n&&n){
&&if(flag)
&&&cout&&尼玛的鸟题目。。害我wa2次，要吸取教训啊。。
&&flag=1;&&int
&&for(i=0;i&n;i++){
&&&cin&&a[i];
&&&sum+=a[i];
&&m=sum/n;
&&int sum1=0;
&&for(i=0;i&n;i++){
&&&if(a[i]&m)
&&&&sum1+=(m-a[i]);
&&cout&&sum1&&
&return 0;
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