【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 ;
 }