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

  1. [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 ...

  2. [cerc2012][Gym100624B]20181013

  3. [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 ...

  4. 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 ...

  5. 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 ...

  6. 4063: [Cerc2012]Darts

    4063: [Cerc2012]Darts Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 85  Solved: 53[Submit][Status] ...

  7. 【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}, ...

  8. 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 ...

  9. 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

  1. c++ 反射类型

    来自: 实现代码=== // // Created by lizhen on 2017/9/29. // #ifndef BOOST_ALL_CALLBACKFUNCTION_H #define BO ...

  2. HashMap get()返回值问题

    问题描述:在进行mysql查询必要字段后,需要根据id进行es其它数据字段的查询拼接.使用HashMap以id为key 以查询过来的数据值为value. 代码如下: Map<String,Int ...

  3. Windows 7中安装Solr7

    最新忙里偷闲,研究一下了Lucene.Net,发现操作比较繁琐,同比相似的功能,感觉Solr比较简单,容易使用.不过由于Solr使用的是Java的环境,对于.Net开发的人员来说,还是比较陌生,搭配环 ...

  4. BAT批处理(五)

    批处理程序 一.交互界面设计 没啥说的,看看设计的菜单界面吧:@echo offclstitle 终极多功能修复:menuclscolor 0Aecho.echo ================== ...

  5. QT分析之QApplication的初始化

    原文地址:http://blog.163.com/net_worm/blog/static/1277024192010097430321/ 在开始分析之前交代一下,一是分析的QT在Window平台实现 ...

  6. c++ new 堆 栈

    根据32位的Windows系统默认有2GB的用户空间,则不能new超过2GB的,执行下列代码: ***]; 会出现下面的错误 error C2148: 数组的总大小不得超过 0x7fffffff 字节 ...

  7. /proc/meminfo中meminfo的计算方法

    /proc/meminfo里的可使用内存的计算没有那么简单,并不是简单的free和page cache的加和 free + pagecache 以此为基准 但是需要减去一些内存:首先要减去系统预留的内 ...

  8. [OS] 操作系统-进程线程-经典面试笔试题

    题目转自:http://blog.csdn.net/morewindows/article/details/7392749 ·线程的基本概念.线程的基本状态及状态之间的关系? 线程,有时称为轻量级进程 ...

  9. Git无法删除文件问题:fatal: pathspec 'readme.txt' did not match any files

    在使用Git时,不小心创建了一个不需要的文件,想要删除一个文件时,出现了错误: fatal: pathspec 'readme.txt' did not match any files 原因是新建的这 ...

  10. Dom-直接 /间接选择器

    Dom:Document Object Model的缩写, 把html转换成了文本对象. 1. 直接选择器 1.直接查找 document.getElementById 根据ID获取一个标签 docu ...