Codeforces 989A:A Blend of Springtime
1 second
256 megabytes
standard input
standard output
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
The first and only line of input contains a non-empty string ss consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (|s|≤100|s|≤100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
.BAC.
Yes
AA..CB
No
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
题意:找一个字符串中是否有连续的三个字符,这三个字符都不相等,且不为"."//
//WA了好多发都是因为大意,没仔细读写好的代码,难受
#include <bits/stdc++.h>
using namespace std;
const int maxn=1000;
char ch[maxn];
int main(int argc, char const *argv[])
{
cin>>ch;
int l=strlen(ch);
int flag=0;
for(int i=0;i<l-2;i++)
{
if(ch[i]!=ch[i+1]&&ch[i+1]!=ch[i+2]&&ch[i+2]!=ch[i]&&ch[i]!='.'&&ch[i+1]!='.'&&ch[i+2]!='.')
flag++;
}
if(flag)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
return 0;
}
Codeforces 989A:A Blend of Springtime的更多相关文章
- Codeforces 731C:Socks(并查集)
http://codeforces.com/problemset/problem/731/C 题意:有n只袜子,m天,k个颜色,每个袜子有一个颜色,再给出m天,每天有两只袜子,每只袜子可能不同颜色,问 ...
- Codeforces 747D:Winter Is Coming(贪心)
http://codeforces.com/problemset/problem/747/D 题意:有n天,k次使用冬天轮胎的机会,无限次使用夏天轮胎的机会,如果t<=0必须使用冬轮,其他随意. ...
- Codeforces 747C:Servers(模拟)
http://codeforces.com/problemset/problem/747/C 题意:有n台机器,q个操作.每次操作从ti时间开始,需要ki台机器,花费di的时间.每次选择机器从小到大开 ...
- Codeforces 749D:Leaving Auction(set+二分)
http://codeforces.com/contest/749/problem/D 题意:有几个人在拍卖场竞价,一共有n次喊价,有q个询问,每一个询问有一个num,接下来num个人从这次拍卖中除去 ...
- Codeforces 749B:Parallelogram is Back(计算几何)
http://codeforces.com/problemset/problem/749/B 题意:已知平行四边形三个顶点,求另外一个顶点可能的位置. 思路:用向量来做. #include <c ...
- Codeforces 749C:Voting(暴力模拟)
http://codeforces.com/problemset/problem/749/C 题意:有n个人投票,分为 D 和 R 两派,从1~n的顺序投票,轮到某人投票的时候,他可以将对方的一个人K ...
- Codeforces 746D:Green and Black Tea(乱搞)
http://codeforces.com/contest/746/problem/D 题意:有n杯茶,a杯绿茶,b杯红茶,问怎么摆放才可以让不超过k杯茶连续摆放,如果不能就输出NO. 思路:首先,设 ...
- Codeforces 745C:Hongcow Builds A Nation(并查集)
http://codeforces.com/problemset/problem/744/A 题意:在一个图里面有n个点m条边,还有k个点是受限制的,即不能从一个受限制的点走到另外一个受限制的点(有路 ...
- Codeforces 743D:Chloe and pleasant prizes(树形DP)
http://codeforces.com/problemset/problem/743/D 题意:求最大两个的不相交子树的点权和,如果没有两个不相交子树,那么输出Impossible. 思路:之前好 ...
随机推荐
- centos7 lua安装
yum -y install gcc automake autoconf libtool makeyum install readline-develcurl -R -O http://www.lua ...
- Sql Server查询同一ID 时间较大的一条数据
- web服务器配置实践
1.为linux系统分配IP地址:192.168.X.1/24,客户端XP系统IP地址为:192.168.X.2/24,其主要DNS指定为:192.168.X.1. 2.查询本机是否安装了httpd服 ...
- AISing Programming Contest 2019 Solution
A - Bulletin Board 签到. #include <bits/stdc++.h> using namespace std; int main() { int n, h, w; ...
- linux下高并发网络应用注意事项
本文转自:http://www.blogjava.net/bacoo/archive/2012/06/11/380500.html linux下高并发网络应用注意事项 vi /etc/sysctl.c ...
- BCG控件初步领略
BCGPVisualStudioGUIDemo 这个界面很不错呀,如果能够实现这种效果,能够解决系列问题 画图程序,这种界面非常先进.用于石材大板等非常优秀. email的效果 这种东西如果效果不错, ...
- [参考]用递归的方法获取 字符 对应的 二进制字符串 (C/C++)
将字符转换为16进制字符串.十进制字符串可以参考这里:https://www.cnblogs.com/stxs/p/8846545.html 代码及调试结果 举例:字符'a',查ASCII码表它对应的 ...
- NOIP2016 T4 魔法阵 暴力枚举+前缀和后缀和优化
想把最近几年的NOIP T4都先干掉,就大概差16年的,所以来做一做. 然后这题就浪费了我一整天QAQ...果然还是自己太弱了QAQ 点我看题 还是pa洛谷的... 题意:给m个物品,每个物品有一个不 ...
- python 生成zip压缩包
import zipfile file_name="a.txt" f = zipfile.ZipFile('test.zip','w',zipfile.ZIP_STORED) f. ...
- json.dump()和json.load()
import json,time # save data to json file def store(data): with open('data.json', 'w') as fw: # 将字典转 ...