【杭电oj】1060 - Leftmost Digit（数学好题）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15463    Accepted Submission(s): 6000 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.

Author

Ignatius.L

巧用lg转化，下面是我用mathtype打出的解题过程：

另外注意，对x取整时应该用__int64，用int会越界，用long int时oj会判PE。

代码如下：

#include <stdio.h>
#include <math.h>
int main()
{
	int u;
	int ans;
	double num,a;
	double x;
	scanf("%d",&u);
	while(u--)
	{
		scanf("%lf",&num);
		x=num*log10(num);
		a=pow(10.0,x-(__int64)x);
		ans=int(a);
		printf("%d\n",ans);
	}
	return 0;
}