RXD is a good mathematician.
One day he wants to calculate:
output the answer module 109+7. 
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
看见这个题不可能去正常做，尝试达标找规律，然后找了 n^K的规律
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
ll ksm(ll n, ll k)
{
ll r = 1;
for (; k; k >>= 1)
{
if (k & 1)
r = r * n % mod;
n = n * n % mod;
}
return r;
}
int main()
{
ll x, y, ca = 1;
while (~scanf("%lld%lld", &x, &y))
{
// x大于mod这题就没法做了
x=x%mod; //利用费马小定理
cout << "Case #" << ca++ << ": " << ksm(x, y) << endl;
}
}