HDU ACM 1060 Leftmost Digit [数学题]

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原题描述

Problem Description

Given a positive integer N, you should output the leftmost digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the leftmost digit of N^N.

Sample Input

2
3
4

Sample Output

2
2

Hint

In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.

解题思路

转一个Discuss里面的解析。
Posted by 2015gdutZhuhui at 2015-11-03 00:31:10 on Problem 1060

需要用到科学记数法和对数运算的知识。
我们把num*num的值记作：num*num=a*10^n，其中1< a <10;
那么，通过两边取对数的方法得到log10(num*num)=log10(a)+n,这时0 < log10(a) < 1;
令x=n+log10(a),得到log10(a)=x-n;所以a=10^(x-n)；
n为整数部分，log10(a)为小数部分，由x=n+log10(a),可知(int)x=n；
最终a=10^（x-n）=10^(x-(int)x)！

参考代码

#include <iostream>#include <cmath>using namespace std;int main(){    int t;    while(~scanf("%d",&t))    {        while (t--)        {            unsigned int n;            scanf("%d", &n);            double x = n * log10(n*1.0);            x -= (__int64)x;            int a = pow(10.0, x);            printf("%d\n", a);        }    }}

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