poj1936

本文介绍了一个简单的算法,用于判断一个字符串是否为另一个字符串的子序列。通过遍历两个字符串并比较字符来实现这一功能,适用于多种编程场景。

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All in All
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 31712 Accepted: 13155

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


暴力解决,简单粗暴

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<malloc.h>
#include<math.h>
int main(){
    char a[100005],b[100005];
    while(~scanf("%s",a)){
        scanf("%s",b);
        int lena=strlen(a);
        int lenb=strlen(b);
        int i,j;
        i=0;j=0;
        while(1){
            if(i==lena){
                printf("Yes\n");
                break;
            }
            else if(i<lena&&j==lenb){
                printf("No\n");
                break;
            }
            if(a[i]==b[j]){
                i++;
                j++;
            }
            else{
                j++;
            }
        }
    }
    return 0;
}






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