poj 1936 All in All

来源:互联网 发布:两种方法创建数据库 编辑:程序博客网 时间:2024/05/29 03:21
All in All
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 27353 Accepted: 11188

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
    判断a串删除若干字符后是否与b串相同
#include <iostream>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>using namespace std;int main(){    char  a[100001],b[100001];    int i,k;    while(~scanf("%s %s",a,b))    {        int lena=strlen(a);        int lenb=strlen(b);          k=0;        for(i=0;i<lenb;i++)            if(a[k]==b[i])               k++;        if(k==lena)            printf("Yes\n");        else printf("No\n");    }    return 0;}


0 0
原创粉丝点击