【Codeforces Round #376 (Div. 2)】 Codeforces 731B Coupons and Discounts

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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).

从左向右扫描，遇到奇数就把后一天减一。更多的减是没有意义的，因为为了保证当前这天是偶数，多减掉的也必须是偶数。如果不能减【后一天是0、已经是最后一天】就是无解。

#include<cstdio>
#include<cstring>
int a[200010];
int main()
{int i,j,k,m,n,p,q,x,y,z;scanf("%d",&n);for (i=1;i<=n;i++)scanf("%d",&a[i]);for (i=1;i<=n;i++)if (a[i]&1){if (i==n||a[i+1]==0){printf("NO\n");return 0;}else a[i+1]--;}printf("YES\n");
}