计算阶乘|集1中的数字-yiteyi-C++库

给定一个整数n，求其阶乘中出现的位数，其中阶乘定义为，阶乘（n）=1*2*3*4*n和阶乘（0）=1 例如：

null

Input :  n = 1Output : 11! = 1 , hence number of digits is 1Input :  5Output : 35! = 120, i.e., 3 digitsInput : 10Output : 710! = 3628800, i.e., 7 digits

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

一个简单的解决方案是计算n！首先，然后计算其中的位数。但是作为n的值！可能非常大，将它们存储在变量中会变得很麻烦（除非您使用的是python！）。更好的解决方案是使用对数的有用特性来计算所需答案。

We know,log(a*b) = log(a) + log(b)Thereforelog( n! ) = log(1*2*3....... * n)           = log(1) + log(2) + ........ +log(n)Now, observe that the floor value of log base 10 increased by 1, of any number, gives thenumber of digits present in that number.Hence, output would be : floor(log(n!)) + 1.

下面是一个相同实施的示例。

C++

                         // A C++ program to find the number of digits in                       
                         // a factorial                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // This function receives an integer n, and returns                       
                         // the number of digits present in n!                       
                         int                                     findDigits(                                     int                                     n)                       
                         {                       
                                                 // factorial exists only for n>=0                       
                                                 if                                     (n < 0)                       
                                                 return                                     0;                       
           
                                                 // base case                       
                                                 if                                     (n <= 1)                       
                                                 return                                     1;                       
           
                                                 // else iterate through n and calculate the                       
                                                 // value                       
                                                 double                                     digits = 0;                       
                                                 for                                     (                                     int                                     i=2; i<=n; i++)                       
                                                 digits +=                                     log10                                     (i);                       
           
                                                 return                                     floor                                     (digits) + 1;                       
                         }                       
           
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 cout << findDigits(1) << endl;                       
                                                 cout << findDigits(5) << endl;                       
                                                 cout << findDigits(10) << endl;                       
                                                 cout << findDigits(120) << endl;                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to find the number                       
                         // of digits in a factorial                       
           
                         import                                     java.io.*;                       
                         import                                     java.util.*;                       
           
                         class                                     GFG                       
                         {                       
                                                 // returns the number of digits                       
                                                 // present in n!                       
                                                 static                                     int                                     findDigits(                                     int                                     n)                       
                                                 {                       
                                                 // factorial exists only for n>=0                       
                                                 if                                     (n <                                     0                                     )                       
                                                 return                                     0                                     ;                       
                                   
                                                 // base case                       
                                                 if                                     (n <=                                     1                                     )                       
                                                 return                                     1                                     ;                       
                                   
                                                 // else iterate through n and calculate the                       
                                                 // value                       
                                                 double                                     digits =                                     0                                     ;                       
                                                 for                                     (                                     int                                     i=                                     2                                     ; i<=n; i++)                       
                                                 digits += Math.log10(i);                       
                                   
                                                 return                                     (                                     int                                     )(Math.floor(digits)) +                                     1                                     ;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main (String[] args)                       
                                                 {                       
                                                 System.out.println(findDigits(                                     1                                     ));                       
                                                 System.out.println(findDigits(                                     5                                     ));                       
                                                 System.out.println(findDigits(                                     10                                     ));                       
                                                 System.out.println(findDigits(                                     120                                     ));                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Pramod Kumar                       

Python3

                         # Python3 program to find the                       
                         # number of digits in a factorial                       
                         import                                     math                       
           
                         # This function receives an integer                       
                         # n, and returns the number of                       
                         # digits present in n!                       
           
                         def                                     findDigits(n):                       
                                   
                                                 # factorial exists only for n>=0                       
                                                 if                                     (n <                                     0                                     ):                       
                                                 return                                     0                                     ;                       
           
                                                 # base case                       
                                                 if                                     (n <                                     =                                     1                                     ):                       
                                                 return                                     1                                     ;                       
           
                                                 # else iterate through n and                       
                                                 # calculate the value                       
                                                 digits                                     =                                     0                                     ;                       
                                                 for                                     i                                     in                                     range                                     (                                     2                                     , n                                     +                                     1                                     ):                       
                                                 digits                                     +                                     =                                     math.log10(i);                       
           
                                                 return                                     math.floor(digits)                                     +                                     1                                     ;                       
           
                         # Driver code                       
                         print                                     (findDigits(                                     1                                     ));                       
                         print                                     (findDigits(                                     5                                     ));                       
                         print                                     (findDigits(                                     10                                     ));                       
                         print                                     (findDigits(                                     120                                     ));                       
           
                         # This code is contributed by mits                       

C#

                         // A C++ program to find the number                       
                         // of digits in a factorial                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 // This function receives an integer                       
                                                 // n, and returns the number of                       
                                                 // digits present in n!                       
                                                 static                                     int                                     findDigits(                                     int                                     n)                       
                                                 {                       
                                   
                                                 // factorial exists only for n>=0                       
                                                 if                                     (n < 0)                       
                                                 return                                     0;                       
                                   
                                                 // base case                       
                                                 if                                     (n <= 1)                       
                                                 return                                     1;                       
                                   
                                                 // else iterate through n and                       
                                                 // calculate the value                       
                                                 double                                     digits = 0;                       
                                                 for                                     (                                     int                                     i = 2; i <= n; i++)                       
                                                 digits += Math.Log10(i);                       
                                   
                                                 return                                     (                                     int                                     )Math.Floor(digits) + 1;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                                 Console.Write(findDigits(1) +                                     ""                                     );                       
                                                 Console.Write(findDigits(5) +                                     ""                                     );                       
                                                 Console.Write(findDigits(10) +                                     ""                                     );                       
                                                 Console.Write(findDigits(120) +                                     ""                                     );                       
                                                 }                       
                         }                       
           
                         // This code is contributed by                       
                         // Smitha Dinesh Semwal                       

PHP

                         <?php                       
                         // PHP program to find                       
                         // the number of digits                       
                         // in a factorial                       
           
                         // This function receives                       
                         // an integer n, and returns                       
                         // the number of digits present in n!                       
           
                         function                                     findDigits(                                     $n                                     )                       
                         {                       
                                                 // factorial exists only for n>=0                       
                                                 if                                     (                                     $n                                     < 0)                       
                                                 return                                     0;                       
           
                                                 // base case                       
                                                 if                                     (                                     $n                                     <= 1)                       
                                                 return                                     1;                       
           
                                                 // else iterate through n and                       
                                                 // calculate the value                       
                                                 $digits                                     = 0;                       
                                                 for                                     (                                     $i                                     = 2;                                     $i                                     <=                                     $n                                     ;                                     $i                                     ++)                       
                                                 $digits                                     += log10(                                     $i                                     );                       
           
                                                 return                                     floor                                     (                                     $digits                                     ) + 1;                       
                         }                       
           
                         // Driver code                       
                         echo                                     findDigits(1),                                     ""                                     ;                       
                         echo                                     findDigits(5),                                     ""                                     ;                       
                         echo                                     findDigits(10),                                     ""                                     ;                       
                         echo                                     findDigits(120),                                     ""                                     ;                       
           
                         // This code is contributed by Ajit.                       
                         ?>                       

Javascript

                         <script>                       
           
                         // A Javascript program to find the number of digits in                       
                         // a factorial                       
           
           
                         // This function receives an integer n, and returns                       
                         // the number of digits present in n!                       
                         function                                     findDigits(n)                       
                         {                       
                                                 // factorial exists only for n>=0                       
                                                 if                                     (n < 0)                       
                                                 return                                     0;                       
           
                                                 // base case                       
                                                 if                                     (n <= 1)                       
                                                 return                                     1;                       
           
                                                 // else iterate through n and calculate the                       
                                                 // value                       
                                                 let digits = 0;                       
                                                 for                                     (let i=2; i<=n; i++)                       
                                                 digits += Math.log10(i);                       
           
                                                 return                                     Math.floor(digits) + 1;                       
                         }                       
           
                         // Driver code                       
                                   
                                                 document.write(findDigits(1) +                                     "<br>"                                     );                       
                                                 document.write(findDigits(5) +                                     "<br>"                                     );                       
                                                 document.write(findDigits(10) +                                     "<br>"                                     );                       
                                                 document.write(findDigits(120) +                                     "<br>"                                     );                       
                                   
           
                         //This code is contributed by Mayank Tyagi                       
           
                         </script>                       

输出：

在下一集中，我们将看到如何进一步优化我们的方法，并降低同一程序的时间复杂度。本文由 阿什图什·库马尔 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用贡献极客。组织或者把你的文章寄到contribute@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。

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