修改

void change(int i, int j, int delta){
    A[i][j] += delta;
    for(int x = i; x < A.length; x += lowbit(x))
        for(int y = j; y < A[i].length; y += lowbit(y))
            C[x][y] += delta;
}

查詢

int getsum(int i, int j){
   int res = 0;
   for(int x = i; x; x -= lowbit(x))
        
 
for(int y = j; y; y -= lowbit(y))
            res += C[x][y];
   return res;
}

子矩陣 (x1, y1) ~ (x2,y2) 的和，我們只需四個矩陣容斥就可以解決：

Ans = sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2)+sum(x1-1,y1-1)

#include<iostream>
#include<stdio.h>
#include<memory.h>
using namespace std;
//int A[1005][1005];
 

int C[1005][1005];
int n,t,x1,y1,x2,y2;
int lowbit(int x)
{
    return x&(-x);
}

void change(int x,int y,int delta)
{
    //A[i][j]+=delta;
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
            C[i][j]+=delta;
}

int getsum(int x,int y)
{
    int res=0;
    for(int
 
 i=x;i;i-=lowbit(i))
        for(int j=y;j;j-=lowbit(j))
            res+=C[i][j];
    return res;
}

//子矩陣(x1,y1)~(x2,y2)的和，只需要四個矩陣容斥即可
//Ans=sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2)+sum(x1-1,y1-1)
int main()
{
    memset(C,0,sizeof(C));
    scanf("%d%d",&n,&t);
    char op[10];
    while(t--)
    {
        scanf("%s",&op);
        if(op[0]==‘C‘)
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            change(x1,y1,1);
            change(x2+1,y1,1);
            change(x1,y2+1,1);
            change(x2+1,y2+1,1);
        }
        else
        {
            scanf("%d%d",&x1,&y1);
            if(getsum(x1,y1)%2==0) printf("0\n");
            else printf("1\n");
        }
    }
    return 0;
}

#include<iostream>
#include<stdio.h>
#include<memory.h>
using namespace std;
//int A[1005][1005];
int C[1005][1005];
int n,t,x1,y1,x2,y2,add;
int lowbit(int x)
{
    return x&(-x);
}

void change(int x,int y,int delta)
{
    //A[i][j]+=delta;
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
            C[i][j]+=delta;
}

int getsum(int x,int y)
{
    int res=0;
    for(int i=x;i;i-=lowbit(i))
        for(int j=y;j;j-=lowbit(j))
            res+=C[i][j];
    return res;
}

//子矩陣(x1,y1)~(x2,y2)的和，只需要四個矩陣容斥即可
//Ans=sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2)+sum(x1-1,y1-1)
int main()
{
    memset(C,0,sizeof(C));
    scanf("%d%d",&n,&t);
    char op[10];
    while(t--)
    {
        scanf("%s",&op);
        if(op[0]==‘C‘)
        {
            scanf("%d%d%d",&x1,&y1,&add);
            change(x1,y1,add);
        }
        else
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            printf("%d\n",getsum(x2,y2)-getsum(x2,y1-1)-getsum(x1-1,y2)+getsum(x1-1,y1-1));
        }
    }
    return 0;
}

二維樹狀數組