CodeForces 1143A The Doors
签到题
#include <iostream>
using namespace std;
int a[200005];
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int x=a[n-1];
int pos=n-2;
for(int i=n-2;i>=0;i--)
{
if(x!=a[i])
{
pos =i;
break;
}
}
printf("%d\n",pos+1);
return 0;
}
CodeForces 1143A The Doors的更多相关文章
- Doors Breaking and Repairing CodeForces - 1102C (思维)
You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consi ...
- Codeforces Round #549 (Div. 2)A. The Doors
A. The Doors time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- Codeforces Round #531 (Div. 3) C. Doors Breaking and Repairing (博弈)
题意:有\(n\)扇门,你每次可以攻击某个门,使其hp减少\(x\)(\(\le 0\)后就不可修复了),之后警察会修复某个门,使其hp增加\(y\),问你最多可以破坏多少扇门? 题解:首先如果\(x ...
- CodeForces 676D代码 哪里有问题呢?
题目: http://codeforces.com/problemset/problem/676/D code: #include <stdio.h> #define MAXN 1001 ...
- Codeforces Round #354 (Div. 2)-D
D. Theseus and labyrinth 题目链接:http://codeforces.com/contest/676/problem/D Theseus has just arrived t ...
- Codeforces Gym 100803C Shopping 贪心
Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopp ...
- Codeforces Round #426 (Div. 2)【A.枚举,B.思维,C,二分+数学】
A. The Useless Toy time limit per test:1 second memory limit per test:256 megabytes input:standard i ...
- Codeforces Round #549 (Div. 2) 训练实录 (5/6)
The Doors +0 找出输入的01数列里,0或者1先出完的的下标. Nirvana +3 输入n,求1到n的数字,哪个数逐位相乘的积最大,输出最大积. 思路是按位比较,从低到高,依次把小位换成全 ...
- ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) D. The Door Problem 2-SAT
题目链接:http://codeforces.com/contest/776/problem/D D. The Door Problem time limit per test 2 seconds m ...
随机推荐
- 出现Failed to get convolution algorithm的解决方法
当运行卷积神经时出现了问题:Failed to get convolution algorithm. This is probably because cuDNN failed to initiali ...
- Vue技术内幕 出去看看吧 挂载
src\platforms\web\runtime\index.js 挂载 Vue.prototype.$mount = function ( el?: string | Element, hydra ...
- JAVA进阶5
间歇性混吃等死,持续性踌躇满志系列-------------第5天 1.IDEA常用快捷键 2.简单方法的使用 package cn.intcast.day05.demo01; public clas ...
- 设计模式二: 工厂方法(Factory Method)
简介 工厂方法模式是创建型模式的一种, 核心结构有四个角色: 抽象工厂,具体工厂,抽象产品,具体产品; 实现层面上,该模式定义一个创建产品的接口,将实际创建工作推迟到具体工厂类实现, 一个产品对应一个 ...
- mysql 客户端连接报错Illegal mix of collations for operation
服务端用的是utf-8,客户端工具打开表和视图是会报Illegal mix of collations for operation错误,经排查,可以采用以下语句解决 SET character_set ...
- Postgresql/Greenplum中将数字转换为字符串TO_CHAR函数前面会多出一个空格
-- 问题1..Postgresql中将数字转换为字符串前面多出一个空格. SELECT TO_CHAR(, '); -- 解决1.使用如下,参数二前面加上fm就可以去掉空格了,如下: SELECT ...
- Docker打包 Asp.Net Core应用,在CentOS上运行(转)
转载连接:https://www.cnblogs.com/ibeisha/archive/2017/09/09/netcoreondocker.html 本文主要介绍下运用docker虚拟技术打包As ...
- 001 python基础实战
报名了阿里大学的AI,一直没有学习,今天开始正式学习. 今天是第一节,Python的基础编程实战,里面包含两个示例. 一:任务实现文件的批量重命名. 1.创建一个目录 2.程序 #!/usr/bin/ ...
- pyqt pyside 设置窗口关闭时删除自身
pyqt pyside 设置窗口关闭时删除自身 self.setAttribute(QtCore.Qt.WA_DeleteOnClose)
- Windows 修改域用户账户密码
打开powershell as administrator Set-ADAccountPassword -Identity 域用户名 弹出提示框,输入旧密码 弹出提示框,输入新密码,两遍 登出 win ...