4 solutions

  • 1
    @ 2025-8-26 9:36:06
    #include<bits/stdc++.h>
    using namespace std;
    int main(){
        int n=1005;
        int a[n][n];
        int h,w;
        cin>>h>>w;
        for(int i=1;i<=h;i++){
            for(int j=1;j<=w;j++){
                cin>>a[i][j];
            }
        }
        for(int i=1;i<=h;i+=2){
            for(int j=1;j<=w;j+=2){
               int s=a[i][j]+a[i+1][j]+a[i][j+1]+a[i+1][j+1]; 
               cout<<s/4<<" ";
            }
            cout<<endl;
        }
        return 0;
    }
    
    • -1
      @ 2025-8-26 15:01:51
      
      #include<bits/stdc++.h>
      using namespace std;
      int main(){
      	//freopen("image.in","r",stdin);
      	//freopen("image.out","w",stdout);
      	int h,w,c=1,d=1,a[1000][1000],b[100][100];
      	cin>>h>>w;
      	for(int i=1;i<=h;i++){
      		for(int j=1;j<=w;j++){
      			cin>>a[i][j];
      		}
      	}
      	for(int i=1;i<=h;i+=2){
      		for(int j=1;j<=w;j+=2){
      			b[c][d]=(a[i][j]+a[i][j+1]+a[i+1][j]+a[i+1][j+1])/4;
      			cout<<b[c][d]<<" ";
      			d+=1;
      		}
      		c+=1;
      		d=1;
      		cout<<endl;
      	}
          return 0;
      }
      ```
      
      
      ```
      • -1
        @ 2025-8-26 14:55:30
        
        #include<bits/stdc++.h>
        using namespace std;
        int main(){
        	//freopen("image.in","r",stdin);
        	//freopen("image.out","w",stdout);
        	int h,w,a[1000][1000];
        	cin>>h>>w;
        	for(int i=1;i<=h;i++){
        		for(int j=1;j<=w;j++){
        			cin>>a[i][j];
        		}
        	}
        	for(int i=1;i<=h;i+=2){
        		for(int j=1;j<=w;j+=2){
        			cout<<(a[i][j]+a[i][j+1]+a[i+1][j]+a[i+1][j+1])/4<<" ";
        		}
        		cout<<endl;
        	}
            return 0;
        }
        ```
        
        
        ```
        • -1
          @ 2025-8-26 11:42:46
          
          #include <bits/stdc++.h>
          using namespace std;
          int n, m, a[1005][1005];
          bool k;
          int main()
          {
          	cin >> n >> m;
          	for (int i = 1;i <= n;i++)
          	{
          		for (int j = 1;j <= m;j++)
          		{
          			cin >> a[i][j];
          		}
          	}
          	for (int i = 2;i <= n;i += 2)
          	{
          		for (int j = 2;j <= m;j += 2)
          		{
          			cout << (a[i - 1][j - 1] + a[i - 1][j] + a[i][j - 1] + a[i][j]) / 4 << " ";
          		}
          		cout << endl;
          	}
          	return 0;
          }
          
          • 1

          Information

          ID
          1028
          Time
          1000ms
          Memory
          256MiB
          Difficulty
          7
          Tags
          # Submissions
          72
          Accepted
          14
          Uploaded By