poj 1936 All in All

All in All

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 31802 Accepted: 13204

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 subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatter

Sample Output

YesNoYesNo

Source

Ulm Local 2002

提示

题意：

您设计了一种新的加密技术，它通过插入它的字符之间随机生成字符串的聪明方式来编码一个消息。我们将不会详细讨论字符串是如何生成并插入到原始消息中的。为了验证您的方法，有必要写一个程序来检查消息是否真的在字符串中。
给定两个字符串s和t，s通过插入字符是否能转化为t。

思路：

没有。

示例程序

Source CodeProblem: 1936Code Length: 487BMemory: 548KTime: 0MSLanguage: GCCResult: Accepted#include <stdio.h>#include <string.h>int main(){    char ch[100001],ch1[100001];    int k,l;    while(scanf("%s %s",ch,ch1)!=EOF)    {        k=strlen(ch)-1;        l=strlen(ch1)-1;        while(k>=0&&l>=0&&l>=k)        {            if(ch[k]==ch1[l])            {                k--;            }            l--;        }        if(k<0)        {            printf("Yes\n");        }        else        {            printf("No\n");        }    }    return 0;}