4 solutions

  • 0
    @ 2024-4-9 20:44:06

    ``

    #include <iostream>
    #include <stdio.h>
    int main() {
        int a[6][6]={0};
        int b[6];
        int x,y;
        for (int i = 1; i < 6; i++) {
            for (int j = 1; j < 6; j++) {
                scanf("%d", &a[i][j]);
            }
        }
        scanf("%d",&x);
        scanf("%d",&y);
        printf("\n");
    
        for(int i=1;i<6;i++){
            b[i]=a[x][i];
        }
            for (int j = 1; j <6 ; ++j) {
                a[x][j]=a[y][j];
            }
        for (int j = 1; j <6 ; ++j) {
            a[y][j]=b[j];
        }
            for (int i = 1; i < 6; i++) {
                for (int j = 1; j < 6; ++j) {
                    printf("%d ", a[i][j]);
                }
                printf("\n");
            }
            return 0;
    
    }
    
    • 0
      @ 2024-2-20 12:53:09

      偷懒用 string ,并且只交换行下标

      #include <bits/stdc++.h>
      using namespace std;
      
      #define DBG(x) cout << #x << "=" << x << endl
      
      const int N = 5;
      int idxs[] = {-1, 1, 2, 3, 4, 5};
      string g[N + 1];
      
      void solve() {
        int m, n;
      
        for (int i = 1; i <= N; i++)
          getline(cin, g[i]);
        cin >> m >> n;
        swap(idxs[m], idxs[n]);
      
        for (int i = 1; i <= N; i++)
          cout << g[idxs[i]] << endl;
      }
      
      int main() {
        solve();
      
        return 0;
      }
      
      • 0
        @ 2023-6-22 18:56:21
        #include <stdio.h>
        int main(){
            int a[6][6];
           int b[6][6];//临时用来交换变量的中间商
            int m,n;
            for(int i=1;i<=5;i++){
                for(int j=1;j<=5;j++){
                    scanf("%d",&a[i][j]);
                }
            }
            scanf("%d%d",&n,&m);
            
            for(int j=1;j<=5;j++){
                b[n][j]=a[n][j];
                a[n][j]=a[m][j];
                a[m][j]=b[n][j];
            }
        
             for(int i=1;i<=5;i++){
                for(int j=1;j<=5;j++){
                    if(j!=5)  printf("%d ",a[i][j]);
                    else printf("%d",a[i][j]);
                }
                printf("\n");
            }
            
            return 0;
        }
        
        • 0
          @ 2022-3-9 21:47:54
          #include <stdio.h>
          #include <algorithm>
          #include <math.h>
          #include <vector>
          #include <iostream>
          #include <string.h>
          #include <queue>
          using namespace std;
          int n,k;
          int a[10][10];
          int main() {
              for(int i=0; i<5; i++) {
                  for(int j=0; j<5; j++)
                      scanf("%d",&a[i][j]);
              }
              scanf("%d%d",&n,&k);
              for(int i=0; i<5; i++) {
                  for(int j=0; j<5; j++) {
                      if(i==n-1) {
                          if(j==4)
                              printf("%d\n",a[k-1][j]);
                          else
                              printf("%d ",a[k-1][j]);
                      } else if(i==k-1) {
                          if(j==4)
                              printf("%d\n",a[n-1][j]);
                          else
                              printf("%d ",a[n-1][j]);
                      } else {
                          if(j==4)
                              printf("%d\n",a[i][j]);
                          else
                              printf("%d ",a[i][j]);
                      }
                  }
              }
              return 0;
          }
          
          
          • 1

          Information

          ID
          86
          Time
          1000ms
          Memory
          256MiB
          Difficulty
          4
          Tags
          # Submissions
          259
          Accepted
          119
          Uploaded By