HDU 1060 Leftmost Digit .

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5483    Accepted Submission(s): 2080

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

 

第一次睇到呢个题目，以为系高精度求幂，然后取第一位…………摆左好耐都无做过，琴日见到华神http://972169909-qq-com.iteye.com/提到

可以利用科学计数法黎做。

众所周知，科学计数法其实就系一个数n = delta * 10^k

题目可以转化为：n^n = delta * 10^k

 

注意到delta个位数部分其实就系最左面既数字，所以我地只要得到delta就咩都搞掂晒啦。

 

所以集合科学计数法后面为10的幂次呢个特点，我地作如下神变：

 

n^n = delta * 10^k

log(n^n) = log(delta * 10^k)
n * log(n) = log(delta) + k

k = floor(n * log(n))
log(delta) = n * log(n) - k = n * log(n) - floor(n * log(n));

delta = 10^(n * log(n) - floor(n * log(n)))

 

求出delta之后，直接int或者floor取整就ok，log(n^n)之所以化成n*log(n)系因为精度问题，n^n会点大家心照啦{= =+}。

k为10既整数幂次，所以显然k = floor(n * log(n))，呢度一定要用floor而唔系int强制转换因为会出错，至于点解会出错，

原因好简单，就系强制double转换int时候既精度缺失{= =}，依个系由浮点数储存机制造成，原理可以参考百*百科或者

伪基百科，呢度懒得9up{= =+}。

 

下面献上代码：

4222331

2011-07-20 10:22:04

Accepted

1060

0MS

272K

493 B

C++

10SGetEternal{（。）（。）}!

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

int main()
{
 int    T;
 double n;

 scanf("%d", &T);
 while (T--)
 {
  scanf("%lf", &n);

#if 0
  printf("test = %lf\n", n * log10(n) - floor(n * log10(n)));
#endif

  //n = n / pow(10, log10(n));
  printf("%d\n", (int)pow(10.0, n * log10(n) - floor(n * log10(n))));
 }
 return 0;
}

/*
n^n = delta * 10^k
log(n^n) = n * log(n) = log(delta) + k
log(delta) = n * log(n) - k = n * log(n) - floor(n * log(n));
*/

 

多谢收睇{= =*}