【 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;
}
来源:CSDN
作者:私忆一秒钟
链接:https://blog.csdn.net/qq_41879343/article/details/102806974