Programming beauty 2--n! The location of the lowest bit 1 in binary representation

Any number inside the computer is represented by a binary representation, which can be used to quickly determine the position of the lowest bit 1 in the binary representation of the n!.

Solution One:

Divides the binary number of a number by 2, and if the end of the binary number is 0, it can be divisible, otherwise it cannot be divisible.

Therefore, the position of the lowest bit 1 in the binary representation of n! is to find out how many factorization 2 are in n!

The following is code 1:

#include <iostream>using namespace Std;int main (void) {int n,m;m=0;cin>>n;while (n) {n>>=1;m+=n;}    cout<< m+1 <<endl; M is the number of n! factor 2, so the final result is to add 1return 0;}

Solution Two:

N! contains the number of factorization 2, equal to n (the number of 1 in the binary representation of N)

(This rule is self-proof)

Code Listing 2:

#include <iostream>using namespace Std;int main (void) {int n,i,j,m;    m=0;    j=0x1;    cin>>n;    I=n;    while (i) {m+= (i&j);i>>=1;    } cout << n-m+1 <<endl; M is the number of n! factor 2, so the final result is to add 1 return 0;

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Programming beauty 2--n! The location of the lowest bit 1 in binary representation