HDU 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12985    Accepted Submission(s): 4973

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

234

 

Sample Output

22
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.
 

 

解释不好，就不解释了

要了解 log的用法、pow函数的用法

#include<stdio.h>#include<math.h>int main(){int T,a;double m;scanf("%d",&T);while(T--){__int64 N;double a;scanf("%I64d",&N);m=N*log10((double)N)-(__int64)(N*log10((double)N));a=pow(10.0,m);printf("%d\n",(int)a);}return 0;