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
代码语言：javascript复制#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;
}
}
