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).
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.
1 //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）。 2 //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…… 3 #include <iostream> 4 #include <cmath> 5 using namespace std; 6 7 int main() 8 { 9 int i,j,k,T; 10 double ans; 11 int num; 12 cin>>T; 13 while(T--) 14 { 15 cin>>num; 16 ans = log10((double)num); 17 ans=ans-(int)ans; 18 ans=ans*num; 19 ans=ans-(int)ans; 20 num=(int)pow(10.0,ans);//10的小于1的数次方肯定只有一位 21 cout<<num<<endl; 22 } 23 return 0; 24 } 25 //注意：涉及到数学函数最好g++提交，否则CE
