11 条题解

  • 11
    @ 2023-10-22 18:52:44

    哪个逼那么不要脸抄我代码

    点赞!!!

    #include <bits/stdc++.h> 
using namespace std; 
long long a[1010][1010],ans[1010][1010]; 
int main() 
{ 
	int n,m,k,t; 
	cin>>n>>m>>k>>t; 
	for(int i=1;i<=n;i++) 
		for(int j=1;j<=m;j++) 
		{ 
			scanf("%lld",&a[i][j]); 
			ans[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1]; 
		} 
	while(k--) 
	{ 
		long x1,x2,y1,y2,num; 
		scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&num); 
		ans[x2+1][y2+1]+=num; 
		ans[x1][y1]+=num; 
		ans[x1][y2+1]-=num; 
		ans[x2+1][y1]-=num; 
	} 
	for(int i=1;i<=n;i++) 
	{ 
		for(int j=1;j<=m;j++) 
		{ 
			ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; 
			printf("%lld ",ans[i][j]);
		} 
		printf("\n"); 
	} 
	for(int i=1;i<=n;i++) 
		for(int j=1;j<=m;j++) 
		ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; 
	while(t--) 
	{ 
		int x1,x2,y1,y2; 
		scanf("%lld%lld%lld%lld",&x1 ,&y1,&x2,&y2); 
		printf("%lld\n",ans[x2][y2]-ans[x1-1][y2]-ans[x2][y1-1]+ans[x1-1][y1-1]); 
	} 
	return 0; 
}

    哪个逼那么不要脸抄我代码

    点赞!!!

    • 3
      @ 2023-10-29 17:40:11 
      #include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
long long n , m , p , q , x , y , x2 , y2 , z;
long long cf[N][N] , a[N][N] , s[N][N];
int main()
{
	cin >> n >> m >> p >> q; 

	for( int i = 1 ; i <= n ; i++ )
	{ 
		for( int j = 1 ; j <= m ; j++ )
		{ 
			cin >> a[i][j]; 
			cf[i][j] = a[i][j] - a[i-1][j] - a[i][j-1] + a[i-1][j-1]; 
		} 
	}

	while( p-- )
	{
		cin >> x >> y >> x2 >> y2 >> z;
		cf[x][y] += z;
		cf[x][y2+1] -= z;
		cf[x2+1][y] -= z;
		cf[x2+1][y2+1] += z;
	}

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

	while( q-- )
	{
		cin >> x >> y >> x2 >> y2;
		cout << s[x2][y2] - s[x-1][y2] - s[x2][y-1] + s[x-1][y-1] << endl;
	}
	return 0;
}
      • 1
        @ 2026-4-17 20:24:55

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

        const int N = 1e3 + 5; long long n,m,p,q,x,y,xx,yy,z; long long d[N][N],a[N][N],s[N][N];

        int main() { cin>>n>>m>>p>>q;

        for(int i=1;i<=n;i++)
{ 
	for(int j=1;j<=m;j++)
	{ 
		cin>>a[i][j]; 
		d[i][j] = a[i][j] - a[i-1][j] - a[i][j-1] + a[i-1][j-1]; 
	} 
}

while(p--)
{
	cin>>x>>y>>xx>>yy>>z;
	d[x][y] += z;
	d[x][yy+1] -= z;
	d[xx+1][y] -= z;
	d[xx+1][yy+1] += z;
}

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

while(q--)
{
	cin>>x>>y>>xx>>yy;
	cout<<s[xx][yy] - s[x-1][yy] - s[xx][y-1] + s[x-1][y-1]<<endl;
}
return 0;

        }

        • 0
          @ 2026-4-22 19:26:48 
          #include <bits/stdc++.h> 
using namespace std; 
long long a[1010][1010],ans[1010][1010]; 
int main() 
{ 
	int n,m,k,t; 
	cin>>n>>m>>k>>t; 
	for(int i=1;i<=n;i++) 
		for(int j=1;j<=m;j++) 
		{ 
			scanf("%lld",&a[i][j]); 
			ans[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1]; 
		} 
	while(k--) 
	{ 
		long x1,x2,y1,y2,num; 
		scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&num); 
		ans[x2+1][y2+1]+=num; 
		ans[x1][y1]+=num; 
		ans[x1][y2+1]-=num; 
		ans[x2+1][y1]-=num; 
	} 
	for(int i=1;i<=n;i++) 
	{ 
		for(int j=1;j<=m;j++) 
		{ 
			ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; 
			printf("%lld ",ans[i][j]);
		} 
		printf("\n"); 
	} 
	for(int i=1;i<=n;i++) 
		for(int j=1;j<=m;j++) 
		ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; 
	while(t--) 
	{ 
		int x1,x2,y1,y2; 
		scanf("%lld%lld%lld%lld",&x1 ,&y1,&x2,&y2); 
		printf("%lld\n",ans[x2][y2]-ans[x1-1][y2]-ans[x2][y1-1]+ans[x1-1][y1-1]); 
	} 
	return 0; 
}
          • 0
            @ 2026-4-22 19:26:23

            #include <bits/stdc++.h> using namespace std; long long a[1010][1010],ans[1010][1010]; int main() { int n,m,k,t; cin>>n>>m>>k>>t; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { scanf("%lld",&a[i][j]); ans[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1]; } while(k--) { long x1,x2,y1,y2,num; scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&num); ans[x2+1][y2+1]+=num; ans[x1][y1]+=num; ans[x1][y2+1]-=num; ans[x2+1][y1]-=num; } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; printf("%lld ",ans[i][j]); } printf("\n"); } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; while(t--) { int x1,x2,y1,y2; scanf("%lld%lld%lld%lld",&x1 ,&y1,&x2,&y2); printf("%lld\n",ans[x2][y2]-ans[x1-1][y2]-ans[x2][y1-1]+ans[x1-1][y1-1]); } return 0; }

            • 0
              @ 2026-4-11 18:31:58

              #include <bits/stdc++.h> using namespace std; const int N = 1e3 + 10; const int INF = 0x3f3f3f3f; long long n , m , p , q , x , y , x2 , y2 , z; long long cf[N][N] , a[N][N] , s[N][N]; int main() { cin >> n >> m >> p >> q;

              for( int i = 1 ; i <= n ; i++ )
{ 
	for( int j = 1 ; j <= m ; j++ )
	{ 
		cin >> a[i][j]; 
		cf[i][j] = a[i][j] - a[i-1][j] - a[i][j-1] + a[i-1][j-1]; 
	} 
}

while( p-- )
{
	cin >> x >> y >> x2 >> y2 >> z;
	cf[x][y] += z;
	cf[x][y2+1] -= z;
	cf[x2+1][y] -= z;
	cf[x2+1][y2+1] += z;
}

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

while( q-- )
{
	cin >> x >> y >> x2 >> y2;
	cout << s[x2][y2] - s[x-1][y2] - s[x2][y-1] + s[x-1][y-1] << endl;
}
return 0;

              }

              • 0
                @ 2026-4-11 18:31:16

                #include <bits/stdc++.h> using namespace std; long long a[1010][1010],ans[1010][1010]; int main() { int n,m,k,t; cin>>n>>m>>k>>t; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { scanf("%lld",&a[i][j]); ans[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1]; } while(k--) { long x1,x2,y1,y2,num; scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&num); ans[x2+1][y2+1]+=num; ans[x1][y1]+=num; ans[x1][y2+1]-=num; ans[x2+1][y1]-=num; } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; printf("%lld ",ans[i][j]); } printf("\n"); } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; while(t--) { int x1,x2,y1,y2; scanf("%lld%lld%lld%lld",&x1 ,&y1,&x2,&y2); printf("%lld\n",ans[x2][y2]-ans[x1-1][y2]-ans[x2][y1-1]+ans[x1-1][y1-1]); } return 0; }

                • 0
                  @ 2024-10-29 17:09:32 
                  #include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
long long n , m , p , q , x , y , x2 , y2 , z;
long long cf[N][N] , a[N][N] , s[N][N];
int main()
{
	cin >> n >> m >> p >> q; 

	for( int i = 1 ; i <= n ; i++ )
	{ 
		for( int j = 1 ; j <= m ; j++ )
		{ 
			cin >> a[i][j]; 
			cf[i][j] = a[i][j] - a[i-1][j] - a[i][j-1] + a[i-1][j-1]; 
		} 
	}

	while( p-- )
	{
		cin >> x >> y >> x2 >> y2 >> z;
		cf[x][y] += z;
		cf[x][y2+1] -= z;
		cf[x2+1][y] -= z;
		cf[x2+1][y2+1] += z;
	}

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

	while( q-- )
	{
		cin >> x >> y >> x2 >> y2;
		cout << s[x2][y2] - s[x-1][y2] - s[x2][y-1] + s[x-1][y-1] << endl;
	}
	return 0;
}
小鸟专属
                  • 0
                    @ 2023-10-22 20:30:35

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

                    long long n,m,p,q,cf[N][N],a[N][N],s[N][N]; long long x,y,x2,y2,z;

                    int main(){ cin>>n>>m>>p>>q; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ cin>>a[i][j]; cf[i][j]=a[i][j]-a[i][j-1]+a[i-1][j-1]; } }

                    while(p--){
	cin>>x>>y>>x2>>y2>>z;
	cf[x][y]+=z;
	cf[x][y2+1]-=z;
	cf[x2+1][y]-=z;
	cf[x2+1][y2+1]+=z;
}

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

while(q--){
	cin>>x>>y>>x2>>y2;
	cout<<s[x2][y2]-s[x-1][y2]-s[x2][y-1]+s[x-1][y-1]<<endl;
}
return 0;

                    }

                    • 0
                      @ 2023-10-11 20:31:04

                      #include #include #include #include using namespace std; const int N = 1e3 + 10; long long n , m , p , q , a[N][N] , s[N][N] , x , y , x2 , y2 , cf[N][N] , tmp; int main(){ cin >> n >> m >> q; for(int i = 1;i <= n;i++) { for(int j = 1; j <= m;j++) { cin >> a[i][j]; cf[i][j] = a[i][j] - a[i-1][j] - a[i][j-1] + a[i-1][j-1]; } }

                      while(p--)
{
	cin >> x >> y >> x2 >> y2;
	cout << sum[x2][y2] - sum[x-1][y2] - sum[x2][y-1] + sum[x-1][y-1] << endl;
	
}

return 0;

                      }保存一下代码哈，晚点再发真代码

                    • 0
                      @ 2023-4-30 16:54:41

                      #include<bits/stdc++.h> using namespace std; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; long long a[1010][1010],ans[1010][1010]; int main() { int n,m,k,t; cin>>n>>m>>k>>t; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { scanf("%lld",&a[i][j]); ans[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1]; } while(k--) { long long x1,x2,y1,y2,num; scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&num); ans[x2+1][y2+1]+=num; ans[x1][y1]+=num; ans[x1][y2+1]-=num; ans[x2+1][y1]-=num; } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; printf("%lld ",ans[i][j]); } printf("\n"); } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) ans[i][j]+=ans[i][j-1]+ans[i-1][j]-ans[i-1][j-1]; while(t--) { int x1,x2,y1,y2; scanf("%lld%lld%lld%lld",&x1 ,&y1,&x2,&y2); printf("%lld\n",ans[x2][y2]-ans[x1-1][y2]-ans[x2][y1-1]+ans[x1-1][y1-1]); } return 0; }

                      • 1

                      二维差分数组模版

                      信息

                      ID
                      1285
                      时间
                      1000ms
                      内存
                      256MiB
                      难度
                      7
                      标签
                      递交数
                      819
                      已通过
                      180
                      上传者
                      1. Language
                      2. 兼容模式
                      3. 主题
                      3. Worker 0, 29ms
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