hdu 6027 水题注意全程取模，全程long long

最新推荐文章于 2019-05-29 13:52:11 发布

原创最新推荐文章于 2019-05-29 13:52:11 发布 · 476 阅读

0 ·

CC 4.0 BY-SA版权

水题专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种解决大数幂求和问题的算法，并提供了一个具体的实现案例。该问题要求计算从1到n的整数k次幂之和并对结果取模，以避免数值过大。文章中的代码采用C++编写，通过双重循环来完成幂运算和求和过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Easy Summation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 969 Accepted Submission(s): 391

Problem Description

You are encountered with a traditional problem concerning the sums of powers.
Given two integers

n and

k. Let

f(i)=ik, please evaluate the sum

f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of

n in this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo

109+7.

Input

The first line of the input contains an integer

T(1≤T≤20), denoting the number of test cases.
Each of the following

T lines contains two integers

n(1≤n≤10000) and

k(0≤k≤5).

Output

For each test case, print a single line containing an integer modulo

109+7.

Sample Input

Sample Output

33
30
10

代码：

#include<stdio.h>
#include<iostream>
using namespace std;
const int maxn=1e4+10;
const int mod=1e9+7;

int main(){
    long long t,n,k;
    scanf("%lld",&t);
    long long ans;
    while(t--){
        scanf("%lld%lld",&n,&k);
        ans=0;
        for(long long i=1;i<=n;++i){
            long long tmp=1;
            for(long long j=1;j<=k;++j) tmp=tmp*i%mod;//这里没注意到取模 WA了两发
            ans+=tmp;
            ans%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}