light_oj 1282 求n^k的前几位数和后几位数

light_oj 1282 求n^k的前几位数和后几位数

E - Leading and Trailing

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1282

Description

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

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

题意：输出n^k的前三位和后三位。

思路：后三位直接快速幂取模，前三位化为科学计数法取对数推导：

            n^k=a.bc*10^m ( m为n^k的位数，即m=(int)lg(n^k)=(int)(k*lgn) );

            求对数:  k*lgn=lg(a.bc)+m

             即 a.bc=10^(k*lgn-m)=10^(k*lgn-(int)(k*lgn));

              abc=a.bc*100;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);
const int p=1000;

ll n,k;

ll qpow(ll n,ll k)
{
    ll res=1;
    while(k){
        if(k&1) res=(res%p)*(n%p)%p;
        n=(n%p)*(n%p)%p;
        k>>=1;
    }
    return res;
}

ll f(ll n)
{
    double x=k*log10(n)-(int)(k*log10(n));
    return pow(10,x)*100;
}

int main()
{
    int T;cin>>T;
    int tag=1;
    while(T--){
        cin>>n>>k;
        printf("Case %d: %lld %03lld\n",tag++,f(n),qpow(n,k));
    }
    return 0;
}