HDU 1060 Leftmost Digit（求 n ^ n 的最高位的值）

Leftmost Digit

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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. 
         

 

題目大意：

給你一個 n  ，然後求出  n ^ n 的最高位的數值是多少。

對一個數num可寫爲 num=10ⁿ + a, 即科學計數法，使a的整數部分即爲num的最高位數字

num^num=10ⁿ + a 這裏的n與上面的n不等

兩邊取對數： num*lg(num) = n + lg(a);

因爲a<10，所以0<lg(a)<1

令x=n+lg(a); 則n爲x的整數部分，lg(a)爲x的小數部分

又x=num*lg(num);

a=10^(x-n) = 10^(x-int(x)))

再取a的整數部分即得num的最高位

附上代碼：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		double n;
		scanf("%lfd",&n);
		double a = n * log10(n);
		long long int b = (long long int)a;
		double c = a - b;
		long long int ans = (long long int)pow(10,c);
		printf("%I64d\n",ans);
	}
	return 0;
}