RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 293    Accepted Submission(s): 150

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  //題意

：已知n,k，求表示式的值。 //思路：那個μ(n)是莫比烏斯函式，p代表的是質數，具體瞭解自行百度。 這題屬於找規律，算幾個就能發現規律： n^k=2：所求值=2； n^k=3：所求值=3； n^k=4：所求值=4； n^k=5：所求值=5； ...... n^k：所求值n^k； 這題就是求（n^k）%（1e9+7），套個快速冪即可。不過快速冪裡有個細節要注意，程式碼裡已標註。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

typedef long long ll;

const ll MOD = 1e9 + 7;

ll n, k;

ll multiply(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1)
		{
			ans = ((ans%MOD)*(a%MOD)) % MOD;
			b--;
		}
		b /= 2;
		//注意這裡a要先%MOD，不然第一次a*a會超過long long,會WA！！！
		a = ((a%MOD)*(a%MOD)) % MOD;
	}
	return ans;
}

int main()
{
	int Case = 1;
	while (scanf("%lld%lld", &n, &k) != EOF)
	{
		ll sum = multiply(n, k);
		printf("Case #%d: %lld\n", Case++, sum%MOD);
	}
	return 0;
}