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  • (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;
    }
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  • 原文地址:https://www.cnblogs.com/cpoint/p/2063865.html
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