Leading and Trailing LightOJ - 1282

输入n， k   取n的前k位和后k位 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

ll pow(ll x, ll n, ll mod) {
	ll  ans = 1;

	while(n>0) {
		if(n&1) {
			ans *= x;
			ans %= mod;
		}
		x *= x;
		x %= mod;

		n >>= 1;
	}

	return ans;
}
int main() {
	int T;

	cin >> T;

	for(int i=1; i<=T; i++){
		ll n, k;

		cin >> n >> k;
		double x = k*log10(n) - (int)(k*log10(n));//取小数部分 
	
		printf("Case %d: %d %03lld\n", i, (int) (pow(10, x)*100), pow(n, k, 1000));
	}
	return 0;
}