杭电1060 Leftmost Digit

解析复杂数值运算与对数求解技巧

最新推荐文章于 2016-04-17 15:21:17 发布

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最新推荐文章于 2016-04-17 15:21:17 发布

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Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15566 Accepted Submission(s): 6048

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

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 is2
       
       
        这个n^n很大，所以考虑用对数求，假设a为n^n的最左为，n^n=a*10^m,两边取对数n*lgn=m+lga;
       
       
        a=10^(n*lgn-m),可以看出m为n*lgn的整数部分，所以可以使用强制转由你n*lgn得到m
       
       
       #include<stdio.h>
#include<math.h>
int main()
{
	int a,t;
	long long n;
	double s;
	scanf("%d",&t);
while(t--)
{  
   scanf("%lld",&n);	
	s=n*log10(n);
	a=1*pow(10,s-(long long)s);
	printf("%d\n",a);
}
	return 0; 
}

Author

Ignatius.L

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