HDOJ-1060-Leftmost Digit（求n^n的最高位）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2924    Accepted Submission(s): 1026

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.
对一个数num可写为 num=10n + a,  即科学计数法，使a的整数部分即为num的最高位数字

numnum=10n + a  这里的n与上面的n不等

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

又x=num*lg(num);

a=10(x-n)  =  10(x-int(x)))

再取a的整数部分即得num的最高位
#include <iostream>#include <math.h>using namespace std;int main(){    int t;    while(cin>>t)    {        while(t--)        {            unsigned long n;            cin>>n;            double x=n*log10(n*1.0);            x-=(__int64)x;            int a=pow(10.0, x);            cout<<a<<endl;        }    }    return 0;}