Leading and Trailing [数学]

方法：a^k=10^(lga^k)=10^k(⋅lga)=10^([k⋅lga]+{k⋅lga})

前n位：10^(n−1)⋅10^{k⋅lga}

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.

S

#include <stdio.h>
#include <math.h>
#include <algorithm>
typedef long long ll;
using namespace std;
ll _pow(ll a,ll n,ll mod)
{
	ll ans=1;
	while(n)
	{
		if(n&1) ans=ans*a%mod;
		a=a*a%mod;
		n>>=1;
	}
	return ans;
}
int main()
{
	int t;
	ll k,n;
	scanf("%d",&t);
	for(int i=1;i<=t;i++)
	{
		scanf("%lld%lld",&n,&k);
		double a=(double)k*log10(n)-floor((double)k*log10(n));//floor和int都是向下取整，floor返回的是浮点型 
		a=pow(10.0,a);
		a=floor(100*a);
		ll b=_pow(n,k,1000);
		printf("Case %d: %0.0lf %0.3lld\n",i,a,b);
	}
	return 0;
	
 } 

ample 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