博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
poj1936——All in All
阅读量:2343 次
发布时间:2019-05-10

本文共 1455 字,大约阅读时间需要 4 分钟。

Description

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.

Output

For each test case output “Yes”, if s is a subsequence of t,otherwise output “No”.

Sample Input

sequence subsequence

person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter
Sample Output

Yes

No
Yes
No

给两个串,判断第一个串是不是第二个串的子序列

#include 
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f#define MAXN 100005#define Mod 10001using namespace std;int main(){ char s[MAXN],t[MAXN]; while(cin>>s>>t) { int i,j; for(i=0,j=0;t[j]!='\0';++j) if(s[i]==t[j]) i++; if(s[i]=='\0') printf("Yes\n"); else printf("No\n"); } return 0;}

转载地址:http://dxcvb.baihongyu.com/

你可能感兴趣的文章
Redis持久化的两种方式
查看>>
判断一个数组,是否可以分成两个数组之和相等的数组
查看>>
背包问题
查看>>
结构体变量之间的比较和赋值原理
查看>>
C++ const修饰函数、函数参数、函数返回值
查看>>
将单链表的每k个节点之间逆序
查看>>
删除链表中重复的节点——重复节点不保留
查看>>
2018腾讯校招编程题——最重要的城市
查看>>
删除链表中重复的节点——重复节点保留一个
查看>>
实战c++中的vector系列--正确释放vector的内存(clear(), swap(), shrink_to_fit()).md
查看>>
链表排序.md
查看>>
进程与线程的区别与联系、进程与线程的通信方式
查看>>
C++与C的区别
查看>>
产生死锁的必要条件及处理方法
查看>>
TCP和UDP的区别
查看>>
事务具有四个特性
查看>>
Hadoop Hdfs 配置
查看>>
tsung集群测试
查看>>
oracle定时删除表空间的数据并释放表空间
查看>>
解决文件提示: /bin/ksh^M: bad interpreter: bad interpreter:No such file or directory
查看>>