hihocoder-Week200-Shorteniring Sequence

hihocoder-Week200-Shorteniring Sequence

题目1 : Shortening Sequence

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

There is an integer array A1, A2 ...AN. Each round you may choose two adjacent integers. If their sum is an odd number, the two adjacent integers can be deleted.

Can you work out the minimum length of the final array after elaborate deletions?

输入

The first line contains one integer N, indicating the length of the initial array.

The second line contains N integers, indicating A1, A2 ...AN.

For 30% of the data：1 ≤ N ≤ 10

For 60% of the data：1 ≤ N ≤ 1000

For 100% of the data：1 ≤ N ≤ 1000000, 0 ≤ Ai ≤ 1000000000

输出

One line with an integer indicating the minimum length of the final array.

样例提示

(1,2) (3,4) (4,5) are deleted.

样例输入

7
1 1 2 3 4 4 5

样例输出

1

简单的dp问题。

已知长度为n的数组删除元素到最后，一定是两种类型，全是偶数，or 全是奇数。

全是偶数，那么n+1位置添加进来一个奇数，则长度减一，否则加一。

找出这条规律，可以进行dp迭代操作。

#include <cstdio> 
#include <cstdlib> 
const int MAXN = 1000000 + 10; 

#define fabs(a) (a)>0?(a):(-a) 

int n, ans, num[MAXN], dp[MAXN]; 

int main(){
	freopen("in.txt", "r", stdin); 

	while(scanf("%d", &n) != EOF)
	{
		for(int i=0; i<n; ++i )
		{
			scanf("%d", &num[i]); 
		} 

		for(int i=0; i<n; ++i)
		{
			if(i == 0)
			{
				if(num[i] %2 == 0)
				{
					dp[i] = 1; 
				}
				else
				{
					dp[i] = -1; 
				}
			}else{
				if(num[i]%2 == 0)
				{
					dp[i] = dp[i-1] + 1; 
				}
				else
				{
					dp[i] = dp[i-1] - 1; 
				}
			}
		}

		ans = fabs( dp[n-1] ); 

		printf("%d
",  ans );
	}
	return 0; 
}