csu 1556(快速幂)

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/liyinggang/p/5792917.html

本文介绍了一道关于数学序列求和的问题——Jerry's Trouble，并提供了使用快速幂算法求解该问题的具体实现代码。

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1556: Jerry's trouble

Time Limit: 10 Sec Memory Limit: 256 MB
Submit: 787 Solved: 317
[Submit][Status][Web Board]

Description

Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on the door. The type of the sequence of number is

But Jerry’s mathematics is poor, help him to escape from the room.

Input

There are some cases (about 500). For each case, there are two integer numbers n, m describe as above ( 1 <= n < 1 000 000, 1 <= m < 1000).

Output

For each case, you program will output the answer of the sum of the sequence of number (mod 1e9+7).

Sample Input

Sample Output

10
15
30
55
100

题意:求解 (1^m+2^m+...+n^m)%mod
题解:快速幂

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1000000007;
LL pow_mod(LL a,LL n){
    LL ans = 1;
    while(n){
        if(n&1) ans = ans*a%mod;
        a = a*a%mod;
        n>>=1;
    }
    return ans;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        LL res = 0;
        for(int i=1;i<=n;i++){
            res = (res+pow_mod(i,m))%mod;
        }
        printf("%lld\n",res);
    }
}