Easy Summation
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
3
2 5
4 2
4 1
Sample Output
33
30
10
就是给出一个n,和指数
1到n 分别求指数次幂,相加求和
自己错误原因:数据类型
不用快速幂也可
这里幂最大才是5
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
using namespace std;
typedef unsigned long long ll;
ll mod = 1e9+7;
ll quick(ll a, int b){
ll ans = 1;
while(b){
if(b&1)
ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
int main(){
int n;
int a,b;
scanf("%d",&n);
while(n--){
scanf("%d %d",&a,&b);
ll sum = 1;
for(int i=2; i<=a; i++)
sum = (sum + quick(i,b) + mod ) % mod;
printf("%lld\n",sum);
}
return 0;
}
//虽然输入的是int范围内,但是在快速幂的运算中 a = a * a % mod;
//可以仅quick a为ll