[cerc2012][Gym100624C]20181013


题意:用元素符号表示字符串
题解:签到题 简单dp
难点在于把元素符号都改成小写qaq
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std; const int N=,M=; char p[][]={"h","he","li","be","b","c","n","o","f","ne","na","mg","al","si","p","s","cl","ar","k","ca","sc","ti","v","cr","mn","fe","co","ni","cu","zn","ga","ge","as","se","br","kr","rb","sr","y","zr","nb","mo","tc","ru","rh","pd","ag","cd","in","sn","sb","te","i","xe","cs","ba","hf","ta","w","re","os","ir","pt","au","hg","tl","pb","bi","po","at","rn","fr","ra","rf","db","sg","bh","hs","mt","ds","rg","cn","fl","lv","la","ce","pr","nd","pm","sm","eu","gd","tb","dy","ho","er","tm","yb","lu","ac","th","pa","u","np","pu","am","cm","bk","cf","es","fm","md","no","lr"};
char s[N];
int len[M];
bool f[N]; int main()
{
//freopen("a.in","r",stdin);
int pl=;
for(int i=;i<pl;i++) len[i]=strlen(p[i]);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s",s+);
int sl=strlen(s+);
memset(f,,sizeof(f));
f[]=;
for(int i=;i<=sl;i++)
{
for(int j=;j<pl;j++)
{
if(len[j]==)
f[i]=(f[i] || (f[i-] && s[i]==p[j][]));
else if(i>=)
f[i]=(f[i] || (f[i-] && s[i-]==p[j][] && s[i]==p[j][]));
}
}
if(f[sl]) printf("YES\n");
else printf("NO\n");
}
return ;
}
[cerc2012][Gym100624C]20181013的更多相关文章
- [cerc2012][Gym100624D]20181013
题意:一个序列,如果存在一个连续子序列,满足该子序列中没有只存在一次的序列,则原序列为boring,否则non-boring 题解: 分治递归 对一个序列,如果找到了一个只出现一次的数位于a[x],则 ...
- [cerc2012][Gym100624B]20181013
- [cerc2012][Gym100624A]20181013
A 题意:n(n<=20)个国家,每个国家之间有一些债务关系,总体为负债的国家会破产,破产国家的债务关系全部消除.问哪些国家可能成为最后一个唯一存在的国家. 题解: 对于每一个状态,面对若干个负 ...
- BZOJ 4057: [Cerc2012]Kingdoms( 状压dp )
状压dp.... 我已开始用递归结果就 TLE 了... 不科学啊...我dp基本上都是用递归的..我只好改成递推 , 刷表法 将全部公司用二进制表示 , 压成一个数 . 0 表示破产 , 1 表示没 ...
- BZOJ 4059: [Cerc2012]Non-boring sequences ( )
要快速在一段子序列中判断一个元素是否只出现一次 , 我们可以预处理出每个元素左边和右边最近的相同元素的位置 , 这样就可以 O( 1 ) 判断. 考虑一段序列 [ l , r ] , 假如我们找到了序 ...
- 4063: [Cerc2012]Darts
4063: [Cerc2012]Darts Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 85 Solved: 53[Submit][Status] ...
- 【BZOJ4061】[Cerc2012]Farm and factory(最短路,构造)
[BZOJ4061][Cerc2012]Farm and factory(最短路,构造) 题面 BZOJ 然而权限题QwQ. 题解 先求出所有点到达\(1,2\)的最短路,不妨记为\(d_{u,1}, ...
- 2018-10-13 21:30:51 conversion of number systems
2018-10-13 21:30:51 c language 二进制.八进制和十六进制: 1) 整数部分 十进制整数转换为 N 进制整数采用“除 N 取余,逆序排列”法. 十进制数字 36926 转 ...
- 4525: [Cerc2012]Kingdoms
4525: [Cerc2012]Kingdoms 题意 n个国家,两两之间可能存在欠债或者被欠债的关系,一个国家破产:其支出大于收入.问一个国家能否坚持到最后. 思路 很有意思的一道题. dp[s]表 ...
随机推荐
- 计算器软件实现系列(五)策略模式+asp.net
一 策略模式代码的编写 using System; using System.Collections.Generic; using System.Linq; using System.Web; /// ...
- JavaScript:理解事件、事件处理函数、钩子函数、回调函数
详情请点击 http://www.jianshu.com/p/a0c580ed3432
- 把jar包加入本地maven库内
1首先,在项目的pom.xml文件中加入 <dependency><groupId>taobao-alidayu</groupId> //名字随便取不要跟已有的重 ...
- GetTickCount 和getTickCount
GetTickCount:正常读取时间函数 getTickCount:不知道是什么鬼东东函数 都包含在windows.h中..运行出的结果天壤之别~~~
- [剑指Offer] 65.矩阵中的路径
题目描述 请设计一个函数,用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径.路径可以从矩阵中的任意一个格子开始,每一步可以在矩阵中向左,向右,向上,向下移动一个格子.如果一条路径经过了矩阵中 ...
- SpringBoot Web(SpringMVC)
入门工程: package com.example.demo.controller; import com.example.demo.entity.User; import org.springfra ...
- 【bzoj2502】清理雪道 有上下界最小流
题目描述 滑雪场坐落在FJ省西北部的若干座山上. 从空中鸟瞰,滑雪场可以看作一个有向无环图,每条弧代表一个斜坡(即雪道),弧的方向代表斜坡下降的方向. 你的团队负责每周定时清理雪道.你们拥有一架直升飞 ...
- Go语言【第十三篇】:Go语言递归函数
Go语言递归函数 递归,就是在运行的过程中调用自己,语法格式如下: func recursion() { recursion() /* 函数调用自身 */ } func main() { recurs ...
- 优先队列实现 大小根堆 解决top k 问题
摘于:http://my.oschina.net/leejun2005/blog/135085 目录:[ - ] 1.认识 PriorityQueue 2.应用:求 Top K 大/小 的元素 3 ...
- [CF452E]Three strings
题目大意:给你三个字符串$A,B,C$,令$L=min(|A|,|B|,|C|)$,对每个$i\in[1,L]$,求出符合$A_{[a,a+i)}=B_{[b,b+i)}=C_{[c,c+i)}$的三 ...