杭电6027 Easy Summation

最新推荐文章于 2019-05-29 13:52:11 发布

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最新推荐文章于 2019-05-29 13:52:11 发布

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本文介绍了一道经典的算法题目，要求计算特定形式的大数幂求和，并通过模运算处理极大数值的情况。提供了完整的C语言实现代码，展示了如何高效地进行幂运算及求和。

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Easy Summation

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/131072K (Java/Other)

Total Submission(s) : 10 Accepted Submission(s) : 2

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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

Source

2017中国大学生程序设计竞赛 - 女生专场

想法：看懂题目就简单啦！

关键的理解最后一句话please output the answer modulo 109+7.

代码：

#include<stdio.h>
#include<math.h>
#include<math.h>
#define ws 1000000007
int n,k;
long long po(int i)
{
    int j;
    long long sum1=1;
    for(j=1;j<=k;j++)
    {
        sum1=sum1*i;
        sum1=sum1%ws;
    }
    return sum1;
}
long long a[10010];
int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        int i;
       scanf("%d %d",&n,&k);
       long long as=0;
       for(i=1;i<=n;i++)
       {
           as=as+po(i);
           as=as%ws;
       }
       printf("%lld\n",as);
    }
    return 0;
}