求助，40分

P1319 压缩技术

leave_for @ 2023-10-11 22:19:21

求助，40分

下载的测试点数据检查后也是正常的

#include<iostream>
using namespace std;
int main()
{
    short n=0;
    short temp=0;
    short mtrx[20000];
    short loop=0;
    cin>>n;
    while(temp<n*n)
    {
        cin>>mtrx[loop];
        temp+=mtrx[loop];
        loop++;
    }
    temp=0;
    loop=1;
    for(short i=0;loop<n;i++)
    {
        for(short j=0;j<mtrx[i];j++)
        {
            if(temp==n) {cout<<endl;temp=0;loop++;}
            cout<<i%2;
            temp++;
        }
    }
    return 0;
}

WA测试点数据：\ in:

80 0 10 360 4 211 2 400 2 12 10 16 2 78 2 78 2 78 2 18 2 23 11 24 2 78 2 78 2 78 2 78 2 78 2 28 2 48 2 72 2 78 2 78 2 78 2 516 2 400 2 93 4 40 10 575 2 400 2 38 2 78 2 78 2 78 2 18 2 58 2 78 2 78 2 78 2 78 2 78 2 28 2 48 2 72 2 78 2 78 2 78 2 28 43 445 2 400 2 93 4 40

out:

00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000001100000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000011000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000001100000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000000000001100000000000
00000000000000000110000000000000000000000000000000000000000000000001100000000000
00000000000000000000000000000000000000000000000000000000000001100000000000000000
00000000000000000000000000000000000000000000000000000000000001100000000000000000
00000000000000000000000000000000000000000000000000000000000001100000000000000000
00000000000000000000000000000000000000000000000000000000000001100000000000000000
00000000000111111111111111111111111111111111111111111100000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000001100000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000011000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000011110000000000000000000000000000000000000000

怎么回事啊，有没有人来解答一下

by zhouzihang3 @ 2023-10-11 23:05:47

你这个代码很奇怪，也不知道是不是因为习惯不一样……你要压行你就认真压，能不能不要只压那一行if(temp==n) {cout<<endl;temp=0;loop++;}啊喂（

然后就是输入，我个人建议在输入时制表 （因为我的代码就是这么干的） ，这样比较省脑细胞。这里把mtrx数组改成一个变量，然后从头开始按照题干赋值。

我的代码段

while(m<n*n){
        cin>>x;//你的mtrx
        if(y==1){//y存储下一次输入要赋值的量
            for(int i=m;i<m+x;i++){
                a[i]=1;//答案这里为了方便应该存一维
            }
            y=0;
        }else{
            y=1;
        }
        m+=x;//更改m值
    }

接下来输出，使用一个新变量k记录答案数组遍历到的位置，然后两层循环输出

我的代码段

int k=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cout<<a[k];
            k++;
        }
        cout<<endl;
    }

我给你一些建议：

1.变量名要么照题干写，要么使用英文单词，所有变量只用于一个用途，除非你有万全把握

2.我建议（可采取可不采取）能用变量就 撅 不用数组，因为这些用完就丢，还浪费内存

ok，答疑完毕，睡觉

{{C}_{艹}}

41行，最长的讨论区

by sbno333 @ 2023-10-11 23:15:27

改代码就要用最少的改动获得正确


#include<iostream>
using namespace std;
int main()
{
    short n=0;
    short temp=0;
    short mtrx[20000];
    short loop=0;
    cin>>n;
    while(temp<n*n)
    {
        cin>>mtrx[loop];
        temp+=mtrx[loop];
        loop++;
    }
    temp=0;
    loop=1;
    for(short i=0;loop<=n;i++)//此处用<=而不是=
    {
        //cout<<mtrx[i]<<endl;
        for(short j=0;j<mtrx[i];j++)
        {
            if(temp==n) {cout<<endl;temp=0;loop++;}
            cout<<i%2;
            temp++;
        }
        if(temp==n){cout<<endl;temp=0;loop++;}//最大的问题，程序执行到最后temp=n,loop=n时已经该完了，如果不特判后头就不会判断了，因此死循环
    }
    return 0;
}
``` @[zero_egg](/user/615166) 
@[leave_for](/user/1093564)

by zhouzihang3 @ 2023-10-11 23:16:25

@sbno333 好钓

by zhouzihang3 @ 2023-10-11 23:17:24

@sbno333 Markdown贴错了害得我找了两秒半@在哪

by sbno333 @ 2023-10-11 23:17:48

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