Codeforces round 493 div2 B.Cutting (贪心技巧)

B. Cutting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.

There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.

Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4,1,2,3,4,5,4,4,5,5]$ $\to$ two cuts $\to$ $[4,1|2,3,4,5|4,4,5,5]$. On each segment the number of even elements should be equal to the number of odd elements.

The cost of the cut between $x$ and $y$ numbers is $|x-y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.

Input

First line of the input contains an integer $n$ ($2\le n\le 100$) and an integer $B$ ($1\le B\le 100$) — the number of elements in the sequence and the number of bitcoins you have.

Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1\le a_i\le 100$) — elements of the sequence, which contains the equal number of even and odd numbers

Output

Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.

Examples

Input

Copy

6 4
1 2 5 10 15 20

Output

Copy

1

Input

Copy

4 10
1 3 2 4

Output

Copy

0

Input

Copy

6 100
1 2 3 4 5 6

Output

Copy

2

Note

In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.

In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.

In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1+1=2$ bitcoins.

#include<iostream> #include<cstring> #include<string> #include<vector> #include<cmath> #include<algorithm> using namespace std; #define maxn 1000 int num[10000]; int ans[maxn],cnt=0; /* 题目大意：给定n个数和一个上界， 一次切割的条件：切割后的每个子序列必须维护奇偶数的个数相同， 可见其有数学性质，这个性质不可能因为切割次数的增多而破坏。 那么，， 最小化的指标是什么？是切割次数； 有何限制？切割的代价和不超过上界。 分析原子状态，切割一次，代价加上左右数之差，切割次数加上一。 很容易让人想到背包问题，，，没错，我刚开始就是背包的思维走下去的， 后来发现误入歧途了。。。

一方面n实在是小的可怜，可以枚举出所有可切点的位置。 所以者显然就是个贪心问题了，对枚举出的所有断点， 排序然后遍历一遍贪心选取下即可。

*/ int main(){  int n,b;  cin>>n>>b;  int even=0;  int add=0;  for(int i=1;i<=n;++i)   cin>>num[i];  for(int i=1;i<n;++i)  {   if(num[i]&1)  even++;   else   add++;///   if(even!=add) continue;        ans[cnt++]=abs(num[i+1]-num[i]);  }///  sort(ans,ans+cnt);////  int sum=0,anss=0;///  for(int i=0;i<cnt;++i)  {////   sum+=ans[i];   if(sum<=b)   anss++;  }//////  cout<<anss<<endl;//  return 0; }