UVA-11029-Leading and Trailing-（快速幂，log10函数求大数首末数）

本文链接：https://blog.youkuaiyun.com/sdau20163942/article/details/78389320

本文介绍了一种计算n^k的前三位和后三位数字的方法，并提供了一个使用C++实现的具体示例。该算法利用了对数运算来获取前三位数字，并通过模运算得到后三位数字。

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题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1970

part from the novice programmers, all others know that you can’t exactly represent numbers raised to some high power. For example, the C function pow(125456, 455) can be represented in double data type format, but you won’t get all the digits of the result. However we can get at least some satisfaction if we could know few of the leading and trailing digits. This is the requirement of this problem.

Input

The first line of input will be an integer T<1001, where T represents the number of test cases. Each of the next T lines contains two positive integers, n and k. n will fit in 32 bit integer and k will be less than 10000001.

Output

For each line of input there will be one line of output. It will be of the format LLL…TTT, where LLL represents the first three digits of n^k and TTT represents the last three digits of n^k. You are assured that n^k will contain at least 6 digits.

Sample Input

Output for Sample Input

123456 1

123456 2

123...456

152...936

求n^k的前3位和后3位

n^k = X *10 ^(len-1) //len是整数的长度,X是10整数位置只有一位的n^k值的浮点数表示
两边取对数：

K*log10(N)=(len-1)+log10( X )

double tmp=k*log10(n*1.0); //temp等于左边

tmp=tmp-(long long)tmp;//因为log(x)小于1，而（len-1）为整数，所以temp减去整数部分就只剩下log10（x）

tmp=pow(10.0,tmp); //10^[log10（x）]==x

ans1=(ll)(tmp*100); //x*100整数部分为3位

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
#define M 1000005
typedef long long ll;

int n;
ll powmod(ll n,ll k)//快速幂
{
    ll res=1;
    while(k>0)
    {
        if(k&1)
            res=res*n%1000;
        n=n*n%1000;
        k>>=1;
    }
    return res;
}

int main()
{
    int T;
    ll n,k,ans1,ans2;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&k);
        ans2=powmod(n,k);
        double tmp=k*log10(n*1.0);
        tmp=tmp-(long long)tmp;
        tmp=pow(10.0,tmp);
        ans1=(ll)(tmp*100);
        printf("%lld...%03lld\n",ans1,ans2);
    }
    return 0;
}