Leftmost Digit

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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.

将整数与小数分开的函数，modf(),头文件《math.h》

#include
#include
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;
}