HDU 1060 - Leftmost Digit

Leftmost Digit

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 2

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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.

题意：

  给你一个数字n。输出n的n次的第一位。

题解：

  n^n=n*log10(n)。而10的整数次方数的首位为1，因此要求此题的答案仅仅需求出n*log10(n)的小数部分m。由于0<m<10,所以10^m的整数部分（(int)pow(10,m)）即为答案。

參考代码：

#include<stdio.h>
#include<math.h>
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		double a,c,n;
		long long d,b;
		scanf("%lf",&n);
		a=n*log10(n);
		b=(long long)a;//因为题目数据量的限制。这里的b的强制转换为long long型。int型会wa。
		c=a-b;
		d=(int)pow(10.0,c);
		printf("%lld\n",d);
	}
	return 0;
}