• hdu acmsteps 2.1.8 Leftmost Digit


    Leftmost Digit

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

    Total Submission(s): 1982 Accepted Submission(s): 884

    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.
     
     
     
       1:  #include<iostream>
       2:  #include<cmath>
       3:  using namespace std;
       4:  int main(){
       5:      int t;
       6:      unsigned long long n;
       7:      cin>>t;
       8:      while(t--){
       9:          cin>>n;
      10:          long double x=n*log10(n*1.0);
      11:          x-=unsigned long long int(x);
      12:          int ans=pow(10.0,double(x));
      13:          cout<<ans<<'
    ';
      14:      }
      15:  }

    解释:

    对一个数num可写为 number=10n * a,  即科学计数法,使a的整数部分即为num的最高位数字
           numbernumber=10n * a  这里的n与上面的n不等
           两边取对数: number*log10(number) = n + log10(a);
           因为a<10,所以0<log10(a)<1
           令x=n+log10(a); 则n为x的整数部分,log10(a)为x的小数部分
           又x=number*log10(number);
            a=10(x-n) =  10(x-int(x)))
            再取a的整数部分即得number的最高位.

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  • 原文地址:https://www.cnblogs.com/ZJUT-jiangnan/p/3361649.html
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