Leftmost Digit

点击打开链接

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.

将整数与小数分开的函数，modf(),头文件《math.h》

#include<math.h>#include<stdio.h>int main(){double a,b,c;int i,j,k,ncase,m,n;scanf("%d",&ncase);while(ncase--){scanf("%d",&m);a=m*log10(m);b=modf(a,&c);k=(int)pow(10,b);printf("%d\n",k);}return 0;}