Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) C. p-binary 水题

C. p-binary

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.

Examples

input
24 0
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).

In the second sample case, we can represent 24=(24+1)+(22+1)+(20+1).

In the third sample case, we can represent 24=(24−1)+(22−1)+(22−1)+(22−1). Note that repeated summands are allowed.

In the fourth sample case, we can represent 4=(24−7)+(21−7). Note that the second summand is negative, which is allowed.

In the fifth sample case, no representation is possible.

题意

定义p-binary为2^x+p

现在给你一个数x，和一个p。

问你最少用多少个p-binary能构造出x，如果没有输出-1

题解

转化为：

x = 2^x1 + 2^x2 + ... + 2^xn + n*p

首先我们知道任何数都能用二进制表示，如果p=0的话，肯定是有解的。那么答案最少都是x的2进制1的个数。

另外什么情况无解呢，即x-n*p<0的时候肯定无解，可以更加有优化为x-n*p<n的时候无解。

答案实际上就是n，我们从小到大枚举n，然后check现在的2进制中1的个数是否小于等于n。

代码

#include<bits/stdc++.h>
using namespace std;

int Count(int x){
    int number=0;
    for(;x;x-=x&-x){
        number++;
    }
    return number;
}
int main(){
    int n,p,ans=0;
    scanf("%d%d",&n,&p);
    while(1){
        n-=p;
        ans++;
        int cnt=Count(n);
        if(ans>n){
            cout<<"-1"<<endl;
            return 0;
        }
        if(cnt<=ans){
            cout<<ans<<endl;
            return 0;
        }
    }
}