LightOJ - 1282 - Leading and Trailing(数学技巧，快速幂取余）

链接：

https://vjudge.net/problem/LightOJ-1282

题意：

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.

思路:

后三位快速幂取余，考虑前三位。
\(n^k\)可以表示为\(a*10^m\)即使用科学计数法。
对两边取对数得到\(k*log10(n) = log10(a)+m\)
则x = log10(a)是k*log10(n)的小数部分。
a = pow(10, x).就是科学计数法的前面部分。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>

using namespace std;
typedef long long LL;
const int INF = 1e9;

const int MAXN = 1e6+10;
const int MOD = 1e9+7;

LL n, k;

LL PowMod(LL a, LL b)
{
    LL res = 1;
    while(b)
    {
        if (b&1)
            res = res*a%1000;
        a = a*a%1000;
        b >>= 1;
    }
    return res;
}

int main()
{
    int t, cnt = 0;
    scanf("%d", &t);
    while(t--)
    {
        printf("Case %d:", ++cnt);
        scanf("%lld%lld", &n, &k);
        double v = 1.0*k*log10(n);
        v -= (LL)v;
        LL r1 = (LL)(pow(10, v)*100);
        LL r2 = PowMod(n, k);
        printf(" %lld %03lld\n", r1, r2);
    }
    
    return 0;
}