LightOJ 1282 Leading and Trailing(取n^k的最高三位数字)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const ll mod = 1000;

int T, cases = 0;
ll high, low, n, k;

ll quick_pow(ll a, ll b)
{
ll res = 1, tmp = a % mod;
while(b){
if(b & 1) res = res * tmp % mod;
tmp = tmp * tmp % mod;
b >>= 1;
}
return res % mod;
}

int main()
{
scanf("%d", &T);
while(T--){
scanf("%lld%lld", &n, &k);
low = quick_pow(n, k);
double t = 1.0 * k * log10(n * 1.0);
ll m = (ll)(floor(t));
high = (ll)(pow(10.0, t - m + 2.0));
//high = (ll) pow(10.0, 2.0 + fmod(1.0 * k * log10(n * 1.0), 1));
printf("Case %d: %lld %03lld\n", ++cases, high, low);
}
return 0;
}