b9113fced86a32cad0d8 @ 2024-05-23 20:28:04

by All_Wrong_Answer @ 2024-05-23 20:41:12

别抄

#include <bits/stdc++.h>
using namespace std;
int da[100005];
int main(){
    int x,a,s=0,q=1;
    cin>>x;
    while(1){
        if(x*x==s) break;
        cin>>a;
        for(int i=s+1;i<=s+a;i++){
            if(q==1) da[i]=0;
            if(q==0) da[i]=1;
        }
        q=(q+1)%2;
        s+=a;
    }
    for(int i=1;i<=x;i++){
        for(int j=1;j<=x;j++){
            cout<<da[((i-1)*x)+j];
        }
        cout<<endl;
    }
    return 0;
} 

思路就是先按输入的顺序涂色，再按格式输出

by All_Wrong_Answer @ 2024-05-23 20:42:10

@wenzhuoren

by llamn @ 2024-05-23 20:46:22

1. 你链接放的什么鬼玩意
2. 正确链接
   楼主的码：

   
   #include<bits/stdc++.h>
   /*
   #include<windows.h>
   */
   using namespace std;

int a[205]; bool modle[205][205];

int main(){ int n; cin>>n; int x = 0; for(int i = 0;i < n n;){ cin>>a[x]; i += a[x]; x++; } / for(int i = 0;i < 105;i++){ cout<<a[i]<<" "; } system("pause"); / int sum = 0; int num = 0; bool type = false; for(int i = 0;i < n;i++){ for(int j = 0;j < n;j++){ modle[i][j] = type; sum++; if(sum >= a[num]){ num++; sum = 0; type = !type; } / for(int q = 0;q < n;q++){ for(int w = 0;w < n;w++){ cout<<modle[q][w]; } cout<<endl; } cout<<"sum="<<sum<<endl; cout<<"num="<<num<<endl; cout<<"type="<<type<<endl; cout<<"x="<<j<<endl; cout<<"y="<<i<<endl; system("pause"); */ } } for(int i = 0;i < n;i++){ for(int j = 0;j < n;j++){ cout<<modle[i][j]; } cout<<endl; } return 0; }

3. Hack : `3 0 9`
4. 修改 ：先判断是否转换`type`,再记录输出。
5. [My AC code](https://www.luogu.com.cn/record/159976678)

by All_Wrong_Answer @ 2024-05-23 20:46:49

这又不难，入门很合理啊

by llamn @ 2024-05-23 20:46:55

@wenzhuoren