題目鏈接

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.

輸入

The first line contains 2 integers N

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

輸出

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

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破題，non-negative number denoting the sum modulo ，wa了好幾次

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

using namespace std;
const
 
 int N = 1024;
const LL MOD = 1e9+7;

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 = 0;
    for(int i=x;i>0;i-=lowbit(i))
    for(int j=y;j>0;j-=lowbit(j)){
        ret += sum[i][j];
        ret %= MOD;
    }
    return ret;
}
int main(){
    cin>>n>>m; char str[16];
    memset(sum,0,sizeof(sum));
    int x1,y1,x2,y2,d;
    while(m--){
        scanf("%s",str);
        if(str[0]==‘A‘){
            scanf("%d%d%d",&x1,&y1,&d);
            add(x1+1,y1+1,d);
        }
        else{
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            LL total = query(x2+1,y2+1);
            LL small = query(x1+0,y1+0);
            LL s1 = query(x2+1,y1+0);
            LL s2 = query(x1+0,y2+1);
            printf("%lld\n",((total+small-s1-s2)%MOD+MOD)%MOD);
        }
    }
    return 0;
}

hiho 172周 - 二維樹狀數組模板題