【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：

 1 #include<iostream>

 2 #include<cmath>

 3 using namespace std;

 4 int main()

 5 {

 6     int sum;

 7     while(cin>>sum)

 8     {

 9         while(sum--)

10         {

11             double n;

12             scanf("%lf",&n);

13             double x=n*log10(n*1.0);

14             _int64 y=(_int64)x;

15             double xy=x-y;

16             int temp=(int)pow(10.0,xy);

17             printf("%d\n",temp);

18         }

19     }

20     return 0;

21 }

View Code