HDU 6027 Easy Summation【简单相加||快速幂】
Easy Summation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 965 Accepted Submission(s): 389
Problem Description
You are encountered with a traditional problem concerning the sums of powers.
Given two integersn 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 modulo109+7 .
Given two integers
Can you figure the answer out? Since the answer may be too large, please output the answer modulo
Input
The first line of the input contains an integer T(1≤T≤20) ,
denoting the number of test cases.
Each of the followingT lines
contains two integers n(1≤n≤10000) and k(0≤k≤5) .
Each of the following
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
Source
注意:
fast_pow(ll p, ll n)中p可能爆int需写成long long
#include<iostream> #include<algorithm> #include<cmath> #include<cstdio> #include<cstdlib> #include<queue> #include<map> #include<set> #include<stack> #include<bitset> #include<numeric> #include<vector> #include<string> #include<iterator> #include<cstring> #include<ctime> #include<functional> #define INF 0x3f3f3f3f #define ms(a,b) memset(a,b,sizeof(a)) #define pi 3.14159265358979 #define mod 1000000007 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef pair<int, int> P; typedef long long ll; typedef unsigned long long ull; const int maxn = 55; ll fast_pow(ll p, ll n) { ll ans = 1; while (n) { if (n & 1) ans = (ans*p) % mod; p = (p*p) % mod; n >>= 1; } return ans; } int main() { int t; scanf("%d", &t); while (t--) { ll n, k, as = 0; scanf("%lld%lld", &n, &k); for (int i = 1; i <= n; i++) { as = (as + fast_pow(i, k)) % mod; } printf("%lld\n", as); } }
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