3588. Permutation and binary

In combinatorics , We learned permutation numbers .

from n Take out... Of the different elements m（m<=n） The number of all permutations of elements , It's called from n To take m Number of permutations of , Write it down as p(n,m).

The specific calculation method is p(n,m)=n(n−1)(n−2)……(n−m+1)=n!/(n−m)!（ Regulations 0!=1）.

When n and m Not very young , This permutation number is a relatively large number , such as p(10,5)=30240.

If expressed in binary, it is p(10,5)=30240=(111011000100000)b, in other words , At the back 5 A zero .

Our problem is , Given a permutation number , Calculate the number of consecutive zeros following its binary representation .

Input format

Input contains multiple sets of test data .

Each group of data occupies one row , Contains two integers n,m.

Last act 0 0, End of input , No need to deal with .

Output format

Output one row per group of data , A result , Represents the permutation number p(n,m) The binary representation of is followed by how many consecutive zeros .

Data range

1≤m≤n≤10000,
The input can contain at most 100 Group data .

sample input ：

10 5
6 1
0 0

sample output ：

5
1

Code ：

//  How many are there at the end of binary 0 <==>  Decomposing prime numbers  2 What is the index of 
#include <bits/stdc++.h>
using namespace std;
int n, m;

int f(int n, int p) // n How many factorials are there p Multiply   Find the number 
{
    
    int res = 0;
    while (n)
        res += n / p, n /= p;

    return res;
}

int main()
{
    
    int n, m;
    while (cin >> n >> m, n)
    {
    
        cout << f(n, 2) - f(n - m, 2) << endl;
    }

    return 0;
}