【hdu 1060】【求N^N最低位数字】

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12229    Accepted Submission(s): 4674

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1060

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

 

 

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代码：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int sum;
    while(cin>>sum)
    {
        while(sum--)
        {
            double n;
            scanf("%lf",&n);
            double x=n*log10(n*1.0);
            _int64 y=(_int64)x;
            double xy=x-y;
            int temp=(int)pow(10.0,xy);
            printf("%d\n",temp);
        }
    }
    return 0;
}