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

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.
1≤n,k≤1018

μ(n)=1(n=1)

μ(n)=(−1)k(n=p1p2…pk)

μ(n)=0(otherwise)

p1,p2,p3…pk 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 
#include 
using namespace std;
#define ll long long
#define maxn 1000000007
ll 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