HDU 1016（Leftmost Digit）N^N 最左边的数

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10400    Accepted Submission(s): 3957

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

 

 

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

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        __int64 n;
        scanf("%I64d",&n);
        double tem=n*log10(1.0*n)-(__int64)(n*log10(1.0*n));
        double ans=pow(10.0,tem);
        printf("%d\n",(int)ans);
    }
    return 0;
}


感慨一下log10（）非常好用
10^（a+b）=n^n  
n*log(n)=a+b    
b=n*log10(n)-(__int64)(n*log10(n))
10^b            从而表示第一位  
n=87455时，a=4,b=0.941784644.
10^a=10000.10^b=8.7455
从别人处引用而来