2017杭电多校联赛第三场-RXD and math (hdu6063) 找规律快速幂

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RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 418    Accepted Submission(s): 212
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
2017 Multi-University Training Contest - Team 3

题目大意:给你一个如上的公式,然后给你 n 和 k 的值,询问代入公式得到的结果。

解题思路:我们注意到 n 和 k  的数据范围都是 10^18 ,所以一旦涉及到关于求u(i) 的值的想法都是不可能实现的,所以我们就不能去利用莫比乌斯函数去具体求值,所以这个时候我们就需要去打表看当n 和 k 值取不同值时的结果,结果发现结果就是n^k.

ac代码:
#include <iostream>#include <cstdio>using namespace std;#define ll long long#define maxn 1000000007ll power2(ll a, ll b, ll c)//fast_power{    ll res = 1;    a %= c;    while (b)    {        if (b & 1)            res = (res * a) % c;        a = (a * a) % c;        b >>= 1;    }    return res;}int main(){int cas=1;ll a,b;while(scanf("%lld%lld",&a,&b)!=EOF){printf("Case #%d: %lld\n",cas++,power2(a,b,maxn));}return 0;}

题目链接:点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=6063