5 solutions

  • 0
    @ 2024-7-10 15:45:46
    #include<bits/stdc++.h>
    using namespace std;
    
    struct carpet{
    	int x1;
    	int y1;
    	int x2;
    	int y2;
    }carp[1001];
    
    
    int main(){
    	int n,m;
    	cin>>n>>m;
    	int land[n+1][n+1];
    	for(int i=1;i<=n;i++){
    		for(int j=1;j<=n;j++){
    			land[i][j]=0;
    		}
    	}
    	for(int i=1;i<=m;i++){
    		cin>>carp[i].x1>>carp[i].y1>>carp[i].x2>>carp[i].y2;
    	}
    
    		for(int i=1;i<=m;i++){
    			for(int j=carp[i].x1;j<=carp[i].x2;j++){
    				for(int k=carp[i].y1;k<=carp[i].y2;k++){
    					land[j][k]++;
    				}
    			}
    		}
    	
    	for(int i=1;i<=n;i++){
    		for(int j=1;j<=n;j++){
    			cout<<land[i][j]<<" ";
    		}
    		cout<<endl;
    	}
    	return 0;
    }
    

    哎我就是菜

    • -1
      @ 2024-7-10 16:44:17

      #include <bits/stdc++.h> #include #include<unordered_map> #include #include using namespace std; const int maxn=1e3+10; unordered_map<int,int>mp ;

      struct node{ int i; int very; int like; }movies[2000005]; int cmp(node a,node b){ if(a.very=b.very){ return a.like>b.like; } return a.very>b.very; } int main(){ int n,zry,title,m; cin>>n; for(int i=0;i<n;i++){ cin>>zry; mp[zry]++; } cin>>m; for(int i=0;i<m;i++){ cin>>zry; movies[i].i=i+1; if(mp.count(zry)){ movies[i].very=mp[zry]; } } for(int i=1;i<m;i++){ cin>>title; if(mp.count(title)){ movies[i].like=mp[title]; } } sort(movies,movies+m,cmp); cout<<movies[0].i<<endl; return 0;

      }

      • -1
        @ 2024-7-10 15:44:29

        xzy的题解

        #include <bits/stdc++.h> using namespace std; const int N=1e3+10; int a[N][N];//这是原数组 int b[N][N];//这个是差分数组 int main(){ int n,m; cin>>n>>m; while(m--) { int x1,y1,x2,y2; cin>>x1>>y1>>x2>>y2; for(int i=x1;i<=x2;i++){ b[i][y1]++;//这里就是差分数组加 b[i] [y2+1]--;//这里有个公式

        } } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ b[i][j]+=b[i][j-1]; cout<<b[i][j]<<" "; } cout<<endl; } }

        • -2
          @ 2024-7-10 15:29:49

          ####11

          • -2
            @ 2024-7-10 15:28:59
            #include<bits/stdc++.h>
            using namespace std;
            int m,n,a[1005][1005];
            int main(){
            	cin>>n>>m;	
            	int x1,y1,x2,y2;
            	for(int i=1;i<=m;i++){
            	 	cin>>x1>>y1>>x2>>y2;
            	 	for(int j=x1;j<=x2;j++){
            	 		for(int l=y1;l<=y2;l++)
            				a[j][l]++;
            		}
            	}
            	for(int i=1;i<=n;i++){
            	 	for(int j=1;j<=n;j++)
            	 		cout<<a[i][j]<<" ";
            	 		cout<<endl;
            	 }
            	return 0;
            }
            
            • 1

            Information

            ID
            2419
            Time
            1000ms
            Memory
            128MiB
            Difficulty
            2
            Tags
            # Submissions
            11
            Accepted
            10
            Uploaded By