贪心小算法

Problem

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][4,1,2,3,4,5,4,4,5,5] →→ two cuts →→ [4,1|2,3,4,5|4,4,5,5][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 xx and yy numbers is |x−y||x−y| bitcoins. Find the maximum possible number of cuts that can be made while spending no more than BB bitcoins.

Input

First line of the input contains an integer nn (2≤n≤1002≤n≤100) and an integer BB (1≤B≤1001≤B≤100) — the number of elements in the sequence and the number of bitcoins you have.

Second line contains nn integers: a1a1, a2a2, ..., anan (1≤ai≤1001≤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 BB bitcoins.

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：给你一个数组，和一个数B；把数组分成若干部分，每部分中的数奇数和偶数的个数要想等，每一次分割的代价是分割点附近的两个数的差值的绝对值；问在B的条件下，最多能分割几次。

题解：依次找出每个分割点，并记录所需要消耗的代价，然后从小到大排序，贪心选择。

#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 200;
bool ok(int a[], int left, int right) {
    int x = 0, y = 0;
    for(int i = left; i <= right; i++) {
        if(a[i]%2==0) {
            y++;
        } else {
            x++;
        }
    }
    if(x == y) {
        return true;
    } else {
        return false;
    }
}
bool cmp(int a, int b) {
    return a < b;
}
int main() {
    int n, B;
    while(scanf("%d %d", &n, &B) != EOF) {
        int a[MAXN], t[MAXN];
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        int k = 0;
        for(int i = 2; i < n; i += 2) {
            if(ok(a, 1, i) && ok(a, i+1, n)) {
                t[k++] = abs(a[i+1] - a[i]);
            }
        }
        int ans = 0;
        sort(t, t + k, cmp);
        for(int i = 0; i < k; i++) {
            if(B >= t[i]) {
                ans++;
                B -= t[i];
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}