Cutting

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 Bbitcoins.
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
123456789101112131415161718192021222324
一道很簡單的暴力題，但是一直沒有抓住關鍵。
問題的性質：每個切點之間沒有關系，即某個切點切還是不切不影響其他的切點，需要看出來他們之間的不關聯性。
附ac碼

#include<cstdio> #include<cstring> #include<algorithm> using namespace std;int a[105]; //記錄原數組 int b[105]; //記錄奇數個數的前綴和 int c[105]; //記錄所有的切點 int n,s,tmp; int cnt=0,ans;int main() {while(~scanf("%d%d",&n,&s)){memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(c,0,sizeof(c));ans=0;cnt=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);b[i]=a[i]%2+b[i-1];}if(n%2==0 && b[n]==n/2) //如果奇數個數字或者整個數列中奇數和偶數的個數不相等顯然切不成{for(int i=2;i<n;i+=2){if(b[i]==i/2)c[cnt++]=abs(a[i+1]-a[i]);}sort(c,c+cnt);tmp=0;for(int i=0;i<cnt;i++){if(tmp+c[i]<=s){tmp+=c[i];ans++;}else{break;}}}printf("%d\n",ans);}return 0; }

總結

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