Cutting（CF-998B）

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基础算法——贪心

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博客围绕整数序列分割问题展开，给定含相同奇偶个数的整数序列，在 x 和 y 元素间分割花费 |x - y| 元，有 k 元预算。需找出最多分割次数，思路是统计奇偶个数，将符合条件的花费放入数组，用贪心算法选最小花费。

Problem Description

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] → two cuts → [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≤n≤100) and an integer B (1≤B≤100) — the number of elements in the sequence and the number of bitcoins you have.

Second line contains n integers: a1, a2, ..., an (1≤ai≤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

6 4
1 2 5 10 15 20

Output

1

Input

4 10
1 3 2 4

Output

0

Input

6 100
1 2 3 4 5 6

Output

2

题意：给出 n 个数字，保证奇数个数和偶数个数相等，可以把该序列分割成连续的片段，使得每一段的奇数和偶数个数相等，假设在 x 和 y 元素中间分割，那么花费 |x-y| 块钱，现有k块钱，求最多可以分割的次数

思路：从开头统计奇数和偶数的个数，如果某个位置奇数和偶数个数相等就把当前这个花费放入数组，然后贪心的选择最小的即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 100001
#define MOD 1e9+7
#define E 1e-6
#define LL long long
using namespace std;
int a[105];
int b[105],c[105],x[105];
int main()
{
   int n,m;
   cin>>n>>m;
   for(int i=1;i<=n;i++)
   {
       scanf("%d",&x[i]);

       if(x[i]%2==0)
           a[i]=a[i-1]+1;
       else
           a[i]=a[i-1];
       
       if(x[i]%2!=0)
           b[i]=b[i-1]+1;
       else
           b[i]=b[i-1];
   }

   int k=0;
   for(int i=1;i<=n;i++)
   {
       if(a[i]==b[i]&&i!=n)
        {
            k++;
            c[k]=fabs(x[i+1]-x[i]);
        }

   }
  
  sort(c+1,c+k+1);
  
  int ans=0;
  for(int i=1;i<=k;i++)
  {
      if(m-c[i]<0) 
          break;
      m-=c[i];
      ans++;
  }
  cout<<ans<<endl;
  return 0;
}