【Light-oj】-1282 - Leading and Trailing（数论，快速幂，log，好）

1282 - Leading and Trailing

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of n^k.

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 n^k contains at least six digits.

Sample Input	Output for Sample Input
5 123456 1 123456 2 2 31 2 32 29 8751919	Case 1: 123 456 Case 2: 152 936 Case 3: 214 648 Case 4: 429 296 Case 5: 665 669

题意：求n^k前三位数和后三位。log求前三位，快速幂求后三位

求任意前几位不清楚可以先看求第一位：杭电 1060-求n^n第一位数

求前三位： n^k=a*10^m; a就是n^k写成科学计数法小数部分

两边对10去对数，得k*lg(n)=m+lg(a);

  double x=k*log(10)n; pow(10,x-(LL)x)=a     求出了a再 x100就是前三位

求后三位： 快速幂求n^k前三位即可，和求最后一位一样，只不过对1000取余

注意：后三位有0，用03lld来补，比较方便

AC代码：

<span style="font-size:10px;">#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
LL quack(LL n,LL m)
{
	LL sum=1;
	n=n%1000;
	while(m)
	{
		if(m&1)
			sum=sum*n %1000;
		n=n*n%1000;
		m>>=1;
	}
	return sum;
}
int main()
{
	int u,ca=1;
	scanf("%d",&u);
	while(u--)
	{
		LL n,k;
		scanf("%lld%lld",&n,&k);
		printf("Case %d: ",ca++);
		LL ans1,ans2;
		double x=(double)k*(log10(n));
		x=pow(10,x-(LL)x);
		ans1=x*100;
		printf("%lld ",ans1);
		ans2=quack(n,k);
		printf("%03lld\n",ans2);
	}
	return 0;
}
</span>