【 Codeforces Round #596 [Div. 2] C p-binary】

◇◆丶佛笑我妖孽 提交于 2019-12-01 03:43:26

【 Codeforces Round #596 [Div. 2] C p-binary】


Description

Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2x+p, where x is a non-negative integer.

For example, some −9-binary (“minus nine” binary) numbers are: −8 (minus eight), 7 and 1015 (−8=20−9, 7=24−9, 1015=210−9).

The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what’s the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.

For example, if p=0 we can represent 7 as 20+21+22.

And if p=−9 we can represent 7 as one number (24−9).

Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).

Input

The only line contains two integers n and p (1≤n≤109, −1000≤p≤1000).

Output

If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer −1. Otherwise, print the smallest possible number of summands.

Sample Input

24 0

Sample Output

2

Note

0 -binary numbers are just regular binary powers, thus in the first sample case we can represent 24=(24+0)+(23+0).

AC代码:

#include <iostream>
#include <algorithm>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
 
int fun(int x)
{
    int cnt=0;
    while(x) {x^=x&-x; cnt++;}
    return cnt;
}
 
int main()
{
    SIS;
    int n,p,cnt=0;
    cin >> n >> p;
    while(true)
    {
        n-=p;
        cnt++;
        if(n<=0) break;
        if(fun(n)<=cnt)
        {
            if(n<cnt) break;
            cout << cnt << endl;
            return 0;
        }
    }
    cout << -1 << endl;
    return 0;
}
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