题目链接：
LightOJ 1282 Leading and Trailing
题意：
给出n和k求出n^k的高三位和低三位。
分析：
低三位别忘了%03d输出啊！
高三位取log瞎搞啊！
#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;
}