hdu 6063 RXD and math(快速幂)

RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 568    Accepted Submission(s): 306

Problem Description
RXD is a good mathematician.
One day he wants to calculate:
i=1nkμ2(i)×nki−−−√

output the answer module 109+7.
1n,k1018
μ(n)=1(n=1)

μ(n)=(1)k(n=p1p2pk)

μ(n)=0(otherwise)

p1,p2,p3pk are different prime numbers
Input
There are several test cases, please keep reading until EOF.
There are exact 10000 cases.
For each test case, there are 2 numbers n,k.
 
Output
For each test case, output "Case #x: y", which means the test case number and the answer.
Sample Input
10 10
 
Sample Output
Case #1: 999999937
Source
 
Recommend
liuyiding   |   We have carefully selected several similar problems for you:  6066 6065 6064 6063 6062 

题解:n^k % mod

注意 :n因为非常大所以在开始就要先用 mod 运算一遍

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
using namespace std;
const long long m=1e9+7;
long long n,k;
long long solve(long long a,long long b)
{
    a=a%m;
   long long ans=1;
   while(b)
   {
       if (b&1)ans=(ans*a)%m;
       b>>=1;
       a=(a*a)%m;
   }
   return ans;
}
int main()
{
    int cas=0;
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        printf("Case #%d: %lld
",++cas,solve(n,k));
    }
    return 0;
}
原文地址:https://www.cnblogs.com/stepping/p/7272756.html