LightOJ1282——Leading and Trailing

You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk.
Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.
Each case starts with a line containing two integers: n (2 ≤ n < 231) and k (1 ≤ k ≤ 107).
Output
For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that nk contains at least six digits.
Sample Input
5
123456 1
123456 2
2 31
2 32
29 8751919
Sample Output
Case 1: 123 456
Case 2: 152 936
Case 3: 214 648
Case 4: 429 296
Case 5: 665 669

求n^k的前3位和后3位，后三位好解决，快速幂取模，前三位就又是骚操作了
因为n^k==10^ak 然后ak就是一个小数x+y，整数部分x其实 10^x就是表示这个乘积的位数 小数部分y 10^y才是乘以这个位数才是乘积 比如10^x是10000 10^y是1.2345 乘积是12345
那么我们只要求出y就能求出10^y 从而得到前三位

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int MOD=1000;
ll quick_pow(ll n,ll k){
    ll ans=1;
    while(k){
        if(k%2){
            ans=(ans*n)%MOD;
        }
        k/=2;
        n=(n*n)%MOD;
    }
    return ans;
}
int main(void){
    int t;
    int c=1;
    long long n,k;
    scanf("%d",&t);
    while(t--){
        scanf("%lld%lld",&n,&k);
        long long a=pow(10.0,2.0+fmod((double)k*log10((double)n),1));
        long long b=quick_pow(n,k);
        printf("Case %d: %0*lld %0*lld\n",c++,3,a,3,b);
    }
    return 0;
}