5 solutions

  • 1
    @ 2025-3-30 14:42:16

    直接swap

    #include <bits/stdc++.h>
    #define endl '\n'
    using namespace std;
    typedef pair<int, int> PII;
    using ll = long long;
    using ULL = unsigned long long;
    const int N = 1e7 + 5;
    int n, m, a[10][10];
    inline void solve() {
        for (int i = 1; i <= 5; i++) 
            for (int j = 1; j <= 5; j++) 
                cin >> a[i][j];
        cin >> n >> m;
        for (int i = 1; i <= 5; i++) {
            for (int j = 1; j <= 5; j++) {
                swap(a[n][j],a[m][j]);
            }
        }        
        for (int i = 1; i <= 5; i++) {
            for (int j = 1; j <= 5; j++) {
                 cout << a[i][j] << ' ';
            }
            cout << endl;
        }    
    
    }
    int main() {
        ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        int _ = 1;
        //int _; cin >> _;
        while (_--) solve();
        return 0;
    }
    
    
    • 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
            277
            Accepted
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