2.2 The beauty of programming-Do not be scared by the factorial to [zero count of N factorial], the beauty of Programming

2.2 The beauty of programming-Do not be scared by the factorial to [zero count of N factorial], the beauty of Programming
2.2 The beauty of programming-Do not be scared by the factorial [zero count of N factorial]

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/zero-count-of-N-factorial.html

[Question]

Question 1:‍Given an integer N, the factorial N of N! How many zeros are there at the end? Example: N = 10, N! = 3 628 800, N! There are two zeros at the end.

　　Idea: This is mainly to judge the number of 5 in each number, because after multiplication of 5 and an even number, we can get 10, which is equivalent to adding a 0.

Question 2: N! In binary format.

　　Idea: At first glance, it seems that question 2 has nothing to do with Question 1. However, from another perspective, if the binary system returns 0 after the bitwise 1, then the position of the bitwise 1 is the number of zeros after the bitwise 1, it is the same as problem 1, except that one is a decimal table and the other is a binary representation. Here, in all the numbers less than N, the multiples of 2 contribute a multiples of 0, 4 and then 0, and so on. Because the binary representation is actually based on 2, each occurrence of a 2 will end with a 0, so you only need to find N! The number of factor 2.

[Code]

C ++ Code

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/* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/7/8 */ //----------------------------------------------- // 1. calculate N! How many zeros are there at the end? //----------------------------------------------- /* Solution 1: Calculate the 5 index of I (I = 1, 2, 3. N) factorization */ Int count (int n) { Int ret = 0; Int I, j; For (I = 1; I <= N; I ++) { J = I; While (0 = j % 5) { Ret ++; J/= 5; } } Return ret; } /* Solution 1: optimize the cycle and set the cycle step to 5 */ Int count (int n) { Int ret = 0; Int I, j; // Set the cyclic step to 5. For (I = 5; I <= N; I = I + 5) { J = I; While (0 = j % 5) { Ret ++; J/= 5; } } Return ret; } /* Solution 2 z = [N/5] + [N/(5*5)] + [N/(5*5*5)] ...... */ /* [N/5] indicates the number of 5 in N, and [N/(5*5)] indicates the number of 5 in [N/5 */ /* Z is N! Contains the number of prime numbers 5 */ Int count (int n) { Int ret = 0; While (n) { Ret + = N/5; N/= 5; } Return ret; } //----------------------------------------------- // 2 calculate N! Binary represents the position of the second digit 1. //----------------------------------------------- /* 2 = (10), every occurrence of a 2, 1 forward 1 bits, such as binary 10*10*10*10 = (10000 )*/ /* Equal to N! Add 1 to the number containing prime factor 2 */ /* Z = [N/2] + [N/(2*2)] + [N/(2*2*2)] ...... */ Int lastone (int n) { Int ret = 0; While (n) { N> = 1; Ret + = n; } Return ret; } /* Determine whether n is the power of 2 */ Bool is2n (int n) { Return n> 0 & (0 = (n & (n-1 ))); }

[Reference]

Http://blog.csdn.net/eric43/article/details/7570474

Http://blog.csdn.net/zcsylj/article/details/6393308

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/zero-count-of-N-factorial.html