hdu 1060 - Leftmost Digit（求N^N最高位）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10320    Accepted Submission(s): 3923

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

234

 

Sample Output

22
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.

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借鉴了下别人的思路：

http://blog.csdn.net/lovelyloulou/article/details/5237263

令num^num=10^n *a（科学计数法）
两边取对数： num*lg(num) = n + lg(a);
因为a<10，所以0<lg(a)<1
令x=n+lg(a); 则n为x的整数部分，lg(a)为x的小数部分
a=10(x-n)=10(x-int(x)))
再取a的整数部分即得num的最高位

log(n)：以e为底的对数

log10(n)：以10为底的对数

以a为底的对数用换底公式：log(n)/log(a)

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    int t;    double x;    long long n;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        x=n*log10(n*1.0);        //cout<<x<<endl;        x-=(long long)x;        int a=pow(10.0,x);        printf("%d\n",a);    }    return 0;}