Codeforces1247-C. p-binary（找规律+枚举）

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
```cpp
24 0

output

2

input

24 1

output

3

input

24 -1

output

4

input

4 -7

output

2

input

1 1

output

-1

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.

解题思路：

找规律，先枚举有多少项加起来，然后拆分数位，如果1的个数小于枚举的项数，那就是可行的。

AC代码：

/*
                 `-._:  .:'   `:::  .:\         |\__/|           /::  .:'   `:::  .:.-'
                    \      :          \         | :: |          /         :       /
                     \     ::    .     `-_______/ ::  \_______-'   .      ::   . /
                      |  :   :: ::'  :   :: ::'  ::: ::'      :: ::'  :   :: :  |
                      |     ;::         ;::        ;::         ;::         ;::  |
                      |  .:'   `:::  .:'   `:::  .:' :::  .:'   `:::  .:'   `:  |
                      /     :           :           :           :           :    \
                     /______::_____     ::    .     ::    .     ::   _____._::____\
                                    `----._:: ::'  :   :: ::'  _.----'
                                           `--.       ;::  .--'
                                               `-. .:'  .-'
                                                  \    /
                                                   \  /
                                                    \/
 */


#include 
//#include 
//#include 
//#include 
//#include 
//#include 
//#include 
#define read(x) scanf("%lld",&x)
#define re(n) for(int i = 0 ; i < n ; i ++)
#define rev(n) for(int i = n-1 ; i >= 0 ; i --)
#define fill(x,y) memset(x,y,sizeof(x))
const int N=1e6+10;
const int INF=0x3f3f3f3f;
typedef long long ll;
using namespace std;

int main()
{
    ll n,m,i,j,k,t = 1;
    for(int cas = 1 ; cas <= t ; cas ++)
    {
        cin>>n>>m;
        ll ans = 0;
        if(n <= m)
            cout<<"-1"<<endl;
        else{
            while(1)
            {
                ans ++;
                ll tmp = n-ans*m;
                if(tmp < ans)
                {
                    cout<<"-1"<<endl;
                    return 0;
                }
                ll tmp1 = 0;
                while(tmp)
                {
                    tmp1 += tmp%2;
                    tmp /= 2;
                }
                if(tmp1 <= ans)
                    break;
            }
             cout<<ans<<endl;
        }

        //cout<<"Case "<
    }
    return 0;
}