Codeforces Round #325 (Div. 2) A
1 second
256 megabytes
standard input
standard output
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.
One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).
The University works in such a way that every day it holds exactly n lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).
The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the n pairs she knows if there will be a class at that time or not.
Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.
Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.
Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.
The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of lessons at the university.
The second line contains n numbers ai (0 ≤ ai ≤ 1). Number ai equals 0, if Alena doesn't have the i-th pairs, otherwise it is equal to 1. Numbers a1, a2, ..., an are separated by spaces.
Print a single number — the number of pairs during which Alena stays at the university.
5
0 1 0 1 1
4
7
1 0 1 0 0 1 0
4
1
0
0
In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair.
In the last sample Alena doesn't have a single pair, so she spends all the time at home.
题意:从第一个‘1’ 开始计数 两个‘1’中最多有一个‘0’ 直到最后一个‘1’ 统计这个序列的个数
题解:暴力
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n;
int a[];
int ans;
int main()
{
scanf("%d",&n);
ans=;
int f=-;
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
for(int i=;i<=n;i++)
{
if(a[i]==&&(i-f-)==)
ans++;
if(a[i]==)
{
ans++;
f=i;
} }
cout<<ans<<endl;
return ;
}
Codeforces Round #325 (Div. 2) A的更多相关文章
- Codeforces Round #325 (Div. 2) F. Lizard Era: Beginning meet in the mid
F. Lizard Era: Beginning Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- Codeforces Round #325 (Div. 2) D. Phillip and Trains BFS
D. Phillip and Trains Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/ ...
- Codeforces Round #325 (Div. 2) C. Gennady the Dentist 暴力
C. Gennady the Dentist Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586 ...
- Codeforces Round #325 (Div. 2) A. Alena's Schedule 水题
A. Alena's Schedule Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/pr ...
- Codeforces Round #325 (Div. 2) B. Laurenty and Shop 前缀和
B. Laurenty and Shop Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/p ...
- Codeforces Round #325 (Div. 2) Phillip and Trains dp
原题连接:http://codeforces.com/contest/586/problem/D 题意: 就大家都玩过地铁奔跑这个游戏(我没玩过),然后给你个当前的地铁的状况,让你判断人是否能够出去. ...
- Codeforces Round #325 (Div. 2) Laurenty and Shop 模拟
原题链接:http://codeforces.com/contest/586/problem/B 题意: 大概就是给你一个两行的路,让你寻找一个来回的最短路,并且不能走重复的路. 题解: 就枚举上下选 ...
- Codeforces Round #325 (Div. 2) Alena's Schedule 模拟
原题链接:http://codeforces.com/contest/586/problem/A 题意: 大概就是给你个序列..瞎比让你统计统计什么长度 题解: 就瞎比搞搞就好 代码: #includ ...
- Codeforces Round #325 (Div. 2) D bfs
D. Phillip and Trains time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Codeforces Round #325 (Div. 1) D. Lizard Era: Beginning
折半搜索,先搜索一半的数字,记录第一个人的值,第二个人.第三个人和第一个人的差值,开个map哈希存一下,然后另一半搜完直接根据差值查找前一半的答案. 代码 #include<cstdio> ...
随机推荐
- linux学习笔记二:三种网络配置
本文引用自:https://www.linuxidc.com/Linux/2017-05/144370.htm [linux公社] VMware为我们提供了三种网络工作模式,它们分别是:Bridged ...
- 【c学习-5】
int main(){ //二维数组的应用 int i,j; int a[2][3]; for(i=0;i void myFunction(){ int a[3]; int i; int max; f ...
- javaScript的闭包 js变量作用域
js的闭包 js的变量作用域: var a=90; //定义一个全局变量 function test(){ a=123; //使用外层的 a变量 } test(); document.write(&q ...
- php-5.6.26源代码 - opcode处理器,“乘法opcode”处理器
// opcode处理器 - 运算符怎么执行: “*” 乘法opcode处理器 static int ZEND_FASTCALL ZEND_MUL_SPEC_CONST_CONST_HANDLER(Z ...
- PHP 日期处理函数 date() 、mktime()
一.前言 php是世界上最好的语言! 二.介绍 mktime()函数获取当周\当天\当月 /** * 微程-日期工具函数 week: 当周 day: 当天 month: 当月 * @author 狗蛋 ...
- Android内购订单验证 --- php实现
直接上代码: function googleVerify($sdata,$google_public_key) { $sdata = json_decode($sdata,true); $in_app ...
- RepeatMasker使用
RM是library-based,通过相似性比对来识别重复序列,可以屏蔽序列中转座子重复序列和低复杂度序列(默认将其替换成N).使用数据库Dfam和Repbase. The Dfam database ...
- Ubuntu 16.04上安装并配置Postfix作为只发送SMTP服务器
如果大家已经在使用第三方邮件服务方案发送并收取邮件,则无需运行自己的邮件服务器.然而,如果大家管理一套云服务器,且其中安装的应用需要发送邮件通知,那么运行一套本地只发送SMTP服务器则更为理想. 如何 ...
- android stadio 打开别人的工程 一直在编译中
这是因为,他工程的gradle 配置,在你本地找不到,所以,会去网上下.然后解压,使用.这是一个很漫长的过程. *那么怎么做呢 修改项目工程的gradle/wrapper/gradle-wrapper ...
- Android学习记录(2)—Android中数据库的常见操作
android中数据库操作是非常常见了,我们会经常用到,操作的方法也有很多种形式,这里我就把最常见的两种形式记录下来了,以备以后用到方便查看.我就不写注释和解释了,因为android数据库的操作和其它 ...