C. p-binary
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
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 pp (which may be positive, negative, or zero). To combine their tastes, they invented pp-binary numbers of the form 2x+p2x+p, where xx is a non-negative integer.
For example, some −9−9-binary ("minus nine" binary) numbers are: −8−8 (minus eight), 77 and 10151015 (−8=20−9−8=20−9, 7=24−97=24−9, 1015=210−91015=210−9).
The boys now use pp-binary numbers to represent everything. They now face a problem: given a positive integer nn, what's the smallest number of pp-binary numbers (not necessarily distinct) they need to represent nn as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0p=0 we can represent 77 as 20+21+2220+21+22.
And if p=−9p=−9 we can represent 77 as one number (24−9)(24−9).
Note that negative pp-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers nn and pp (1≤n≤1091≤n≤109, −1000≤p≤1000−1000≤p≤1000).
Output
If it is impossible to represent nn as the sum of any number of pp-binary numbers, print a single integer −1−1. Otherwise, print the smallest possible number of summands.
Examples
input
Copy
24 0
output
Copy
2
input
Copy
24 1
output
Copy
3
input
Copy
24 -1
output
Copy
4
input
Copy
4 -7
output
Copy
2
input
Copy
1 1
output
Copy
-1
Note
00-binary numbers are just regular binary powers, thus in the first sample case we can represent 24=(24+0)+(23+0)24=(24+0)+(23+0).
In the second sample case, we can represent 24=(24+1)+(22+1)+(20+1)24=(24+1)+(22+1)+(20+1).
In the third sample case, we can represent 24=(24−1)+(22−1)+(22−1)+(22−1)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)4=(24−7)+(21−7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
直接枚举答案即可
我们发现如果答案存在 那么答案的值一定会很小 是不可能超过100的
我们枚举之后可以理解为num=n-i*p然后只需要判断num是否可以用i个二进制数组成即可
那么我们可以推出一个范围 i是否在 二进制1的个数和num之间即可
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,p;
scanf("%d%d",&n,&p);
int ans=-1;
for(int i=1;i<=100;i++)
{
int num=n-i*p;
if(num<=0) break;
int cnt=0;
while(num)
{
if(num%2) cnt++;
num/=2;
}
num=n-i*p;
if(cnt<=i&&i<=num)
{
ans=i;
break;
}
}
printf("%d\n",ans);
}
