(HDOJ 1060)Leftmost Digit

 Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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.
 
  Sample code:

   #include<stdio.h>

#include<math.h>
int main()
{
     
int n,a;
     scanf(
"%d",&n);
     
while(n--)
     {
       scanf(
"%d",&a);
      
double m=a*log10(a);
      
long long z=(long long)m;
      
double p=m-z;
      
int r=pow(10,p);
      printf(
"%d\n",r);
     }
    
return 0;
}

 

 

posted @ 2011-05-31 00:00  cpoint  阅读(478)  评论(0编辑  收藏  举报
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