hiho 172周 - 二维树状数组模板题

题目链接

描述

You are given an N × N matrix. At the beginning every element is 0. Write a program supporting 2 operations:

1. Add x y value: Add value to the element A_xy. (Subscripts starts from 0

2. Sum x₁ y₁ x₂ y₁: Return the sum of every element A_xy for x₁ ≤ x ≤ x₂, y₁ ≤ y ≤ y₂.

输入

The first line contains 2 integers N and M, the size of the matrix and the number of operations.

Each of the following M line contains an operation.

1 ≤ N ≤ 1000, 1 ≤ M ≤ 100000

For each Add operation: 0 ≤ x < N, 0 ≤ y < N, -1000000 ≤ value ≤ 1000000

For each Sum operation: 0 ≤ x₁ ≤ x₂ < N, 0 ≤ y₁ ≤ y₂ < N

输出

For each Sum operation output a non-negative number denoting the sum modulo 10⁹+7.

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破题，non-negative number denoting the sum modulo ，wa了好几次

#include <cstdio>
#include <cstring>
#include <iostream>
typedef long long LL;

using namespace std;
const int N = ;
const LL MOD = 1e9+;

int n,m;
LL sum[N][N];
int lowbit(int data){
    return data&(-data);
}

void add(int x,int y,int d){
    for(int i=x;i<=n;i+=lowbit(i))
    for(int j=y;j<=n;j+=lowbit(j)){
        sum[i][j] += d;
        sum[i][j] %= MOD;
    }
}
LL query(int x,int y){
    LL ret = ;
    for(int i=x;i>;i-=lowbit(i))
    for(int j=y;j>;j-=lowbit(j)){
        ret += sum[i][j];
        ret %= MOD;
    }
    return ret;
}
int main(){
    cin>>n>>m; char str[];
    memset(sum,,sizeof(sum));
    int x1,y1,x2,y2,d;
    while(m--){
        scanf("%s",str);
        if(str[]=='A'){
            scanf("%d%d%d",&x1,&y1,&d);
            add(x1+,y1+,d);
        }
        else{
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            LL total = query(x2+,y2+);
            LL small = query(x1+,y1+);
            LL s1 = query(x2+,y1+);
            LL s2 = query(x1+,y2+);
            printf("%lld\n",((total+small-s1-s2)%MOD+MOD)%MOD);
        }
    }
    return ;
}