HDU 1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8550 Accepted Submission(s): 3296

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.

 //x ^ x= 10^(x*lg(x))=10^(整数部分+小数部分) ；则x ^ x的最高位是由小数部分决定的（因为10的整数次幂不会影响最高位，只在最末位加0）。
 //去掉整数部分（floor函数向下去整）得到10^(小数部分)最高位即是所求……
 #include <iostream>
 #include <cmath>
 using namespace std;
 
 int main()
 {
     int i,j,k,T;
     double ans;
     int num;
     cin>>T;
     while(T--)
     {
         cin>>num;
         ans = log10((double)num);
         ans=ans-(int)ans;
         ans=ans*num;
         ans=ans-(int)ans;
         num=(int)pow(10.0,ans);//10的小于1的数次方肯定只有一位 
         cout<<num<<endl;
     }
     return 0;
 }
 //注意：涉及到数学函数最好g++提交，否则CE