Coupons and Discounts
1 second
256 megabytes
standard input
standard output
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.
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.
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).
4
1 2 1 2
YES
3
1 0 1
NO
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.
分析:贪心,模拟;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<ll,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
const int maxn=2e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,a[maxn];
int main()
{
int i,j;
scanf("%d",&n);
rep(i,,n)scanf("%d",&a[i]);
rep(i,,n)
{
if(a[i]&)
{
if(a[i+])a[i+]--;
else return *puts("NO");
}
}
puts("YES");
//system("Pause");
return ;
}
Coupons and Discounts的更多相关文章
- Codeforces 376B. Coupons and Discounts
B. Coupons and Discounts time limit per test 1 second memory limit per test 256 megabytes input stan ...
- Codeforces 731B Coupons and Discounts(贪心)
题目链接 Coupons and Discounts 逐步贪心即可. 若当前位为奇数则当前位的下一位减一,否则不动. #include <bits/stdc++.h> using name ...
- 【50.49%】【codeforces 731B】Coupons and Discounts
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- CodeForces 731B Coupons and Discounts (水题模拟)
题意:有n个队参加CCPC,然后有两种优惠方式,一种是一天买再次,一种是买两天,现在让你判断能不能找到一种方式,使得优惠不剩余. 析:直接模拟,如果本次是奇数,那么就得用第二种,作一个标记,再去计算下 ...
- Codeforces Round #376 (Div. 2) A B C 水 模拟 并查集
A. Night at the Museum time limit per test 1 second memory limit per test 256 megabytes input standa ...
- 【Codeforces】Round #376 (Div. 2)
http://codeforces.com/contest/731 不发题面了,自己点链接 总结一下 考场上 原以为这次要加很多raiting... 但FST狗记邓,只加了58rating 总结一下 ...
- codeforces754D Fedor and coupons
本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/ ...
- codeforces 754D. Fedor and coupons
D. Fedor and coupons time limit per test 4 seconds memory limit per test 256 megabytes input standar ...
- CF 161B Discounts(贪心)
题目链接: 传送门 Discounts time limit per test:3 second memory limit per test:256 megabytes Description ...
随机推荐
- CSS3秘笈复习:第七章
1.边距冲突: 当元素的bottom margin碰到另一个元素的top margin可能会产生一些怪异的计算,浏览器会忽略小的那个值而使用大的值. 2.边距折叠: 假设要在警告框里插入一个标题,并且 ...
- Glusterfs[转]
原文地址:http://support.huawei.com/ecommunity/bbs/10253434.html 1. GlusterFS概述 GlusterFS是Scale-Out存 ...
- 谷歌浏览器js debug
- 浙大 pat 1007题解
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, ...
- delphi怎么实现全选的功能
1. SelectAll 可以实现全选功能 Delphi/Pascal code edit1.SelectAll; // Delphi/Pascal code RichEdit1.SelStart:= ...
- PHP中使用CURL(一)
执行流程: curl_init()初始化 -> curl_setopt()设置变量 -> curl_exec()获取结果 -> curl_close()释放句柄 Get: $ch = ...
- 第1章 初识java----Java简介
1.Java最初的名字是OAK,是咖啡的意思,在1995年被重命名为Java. ●Java编程语言,即语法. ●Java文件格式,即各种文件夹.文件的后缀. ●Java虚拟机(JVM),即处理*.cl ...
- 练习3:修改withdraw 方法
练习目标-使用有返回值的方法:在本练习里,将修改withdraw方法以返回一个布尔值来指示交易是否成功. 任务 1.修改Account类 a.修改deposit 方法返回true(意味所有存款是成功的 ...
- Android任务栈TaskStack
Task:有多个Activity按顺序组成的一个完整的业务逻辑. 任务栈(TaskStack):新增的Activity放入栈中,点击back栈顶Activity从栈中退出. android:nohis ...
- PAT 天梯赛 L2-007 家庭房产
建图+DFS 题目链接:https://www.patest.cn/contests/gplt/L2-007 题解 在热身赛的时候没有做出来,用的并查集的思想,但是敲残了,最后也没整出来.赛后听到别人 ...