Codeforces 731 B. Coupons and Discounts【贪心】

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The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.

Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.

There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).

As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.

Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.

Output

If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).

Examples
input
41 2 1 2
output
YES
input
31 0 1
output
NO

Note

In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.

In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.

题目大意：

一共有N天，对应每天需要购买的披萨数目已知，对应当前天可以有两种操作：

①直接买两个披萨

②买一个披萨，并且对下一天预定一个披萨，那么到第二天的时候，那个披萨就相当于买了。

问能否通过上述两种操作将这N天的披萨数都正好对应的购买。

思路：

1、首先考虑如果第N天预定了披萨，那么N+1天的时候就购买了一个披萨，明显这是我们不想要的结果。那么我们贪心的点从这个角度出发。最后一天不能进行预定操作，那么如果最后一天是一个奇数，那么显然那个多余出来的1，一定是从上一天预定过来的。

2、那么根据这个特性，我们逆序思维，从最后一天开始向前处理，如果当前天的数是一个奇数，那么对应一定从前一天预定了一份披萨到今天，那么a【i-1】-=1；那么在这个向前推的过程中遇到了a【i】<0的时候，一定是NO的情况。

3、那么对应已知向前推，一直推到第一个的时候，判断第一个的奇偶性，如果其实奇数，那么输出NO，否则就是YES、

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int a[2004000];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int flag=0;        for(int i=n-1;i>=0;i--)        {            if(a[i]<0)            {                flag=0;break;            }            if(i!=0)            {                if(a[i]%2==1)                {                    a[i-1]-=1;                }            }            else if(a[i]%2==0)flag=1;            else flag=0;        }        if(flag==1)printf("YES\n");        else printf("NO\n");    }}

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