Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3444 Accepted Submission(s): 1525 

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.
 

 

Author

Ignatius.L

挺有意思的数学题，贴上来

求N^N 的最高位是什么数字

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

num^num=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

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
using namespace std;
 int main()
 {
     int t;
    int n;
    scanf("%d",&t);
     while(t--)
    {
        scanf("%d",&n);
         double temp=n*log10(n*1.0);
         double res=temp-floor(temp);
         printf("%d\n",(int)pow(10.0,res));
     }
    return 0;
     }