hiho 172周 - 二维树状数组模板题
Posted redips
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hiho 172周 - 二维树状数组模板题相关的知识,希望对你有一定的参考价值。
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.
----------------------------------------------------------------------------------------------------------------------
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 = 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; }
以上是关于hiho 172周 - 二维树状数组模板题的主要内容,如果未能解决你的问题,请参考以下文章