Easy Summation

最新推荐文章于 2021-10-13 18:45:58 发布

原创最新推荐文章于 2021-10-13 18:45:58 发布 · 183 阅读

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#Easy Summation

本文探讨了在处理大规模数值时，如何有效地计算幂次求和的问题。通过使用long long类型变量并结合取模运算，避免了数据溢出，确保了算法在面对大数据时的稳定性和准确性。

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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 10 9+7.

Sample Input
3
2 5
4 2
4 1

Sample Output
33
30
10
分析：

由于数据太大 最好用long long类型 最后结果要对1000000007取模

代码总览：

#include <stdio.h>
#define N 1000000007
long long maxx(long long x,long long n)
{
    long long t=1,i;
    for(i=1;i<=n;i++)
    {
        t=t*x;
        t=t%N;//防止数据过大
    }
    return t;   
}
int main()
{
    long long t,a,b,i,sum;
    scanf("%lld",&t);
    while(t--){
        sum=0;
        scanf("%lld %lld",&a,&b);
        for(i=1;i<=a;i++)
        {
            sum=sum+maxx(i,b);
            sum=sum%N;//防止数据过大
        }
        printf("%lld\n",sum);   
    }
    return 0;
}