hiho一下 第172周

题目1 : Matrix Sum

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

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 Axy. (Subscripts starts from 0

2. Sum x1 y1 x2 y1: Return the sum of every element Axy 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.

样例输入

5 8
Add 0 0 1
Sum 0 0 1 1
Add 1 1 1
Sum 0 0 1 1
Add 2 2 1
Add 3 3 1
Add 4 4 -1
Sum 0 0 4 4 

样例输出

1
2
3 

肯定是二维线段树，虽然思路明白，但是不会写，先用树状数组吧。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int N, M, x1, x2, y1, y2, v;
long long BIT2[][];
char str[];
int lowbit(int x) {
    return x & (-x);
}
void add(int x, int y, int val) {
    for (int i = x; i <= N; i += lowbit(i)) {
        for (int j = y; j <= N; j += lowbit(j)) {
            BIT2[i][j] += val;
            BIT2[i][j] %= ;
        }
    }
}
long long sum(int x, int y) {
    long long ret = ;
    for (int i = x; i > ; i -= lowbit(i)) {
        for (int j = y; j > ; j -= lowbit(j)) {
            ret += BIT2[i][j];
            ret %= ;
        }
    }
    return ret;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    scanf("%d%d", &N, &M);
    memset(BIT2, , sizeof(BIT2));
    for (int i = ; i < M; i++) {
        scanf("%s", str);
        if (strcmp(str, "Add") == ) {
            scanf("%d%d%d", &x1, &y1, &v);
            x1++, y1++;
            add(x1, y1, v);
        } else {
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            x2++, y2++;
            long long ans = sum(x2, y2);
            ans = ans - sum(x1, y2) - sum(x2, y1) + sum(x1, y1);
            while (ans < ) {
                ans += ;
            }
            printf("%lld\n", ans);
        }
    }
    return ;
}