LightOJ - 1282 B - Leading and Trailing——快速幂运算

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
ll pow_mod(ll a, ll b, ll c) {
    ll ans = 1;
    a %= c;
    while (b) {
        if (b & 1) ans = ans * a % c;
        a = a * a % c;
        b >>= 1;
    }
    return ans;
}
int main()
{
    int T; scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        ll n, k; scanf("%lld %lld", &n, &k);
        printf("Case %d: %d %03d\n", i, (int)pow(10.0, 2.0 + fmod(log10(1.0 * n) * k, 1.0)), (int)pow_mod(n, k, 1000));
    }
    return 0;
}