POJ 1936 All in All

Posted 2020-09-08 joeylee97

Language: Default All in All Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 32980   Accepted: 13750 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 Source #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<string> using namespace std; typedef long long LL; #define MAXN 100003 /* 求s和t最长相同子串，如果长度等于s输出YES */ char s[MAXN],t[MAXN]; int main() { while(scanf("%s%s",s,t)!=EOF) { int l1 = strlen(s),l2 = strlen(t); if(strstr(t,s)) { printf("Yes\n"); continue; } int i=0,j=0; while(i<l1&&j<l2) { if(s[i]!=t[j]) { j++;//如果不匹配 对t串的下一个字符查找 } else { i++;//找到i字符，找下一个 j++; } } if(i==l1)//说明s串中所有字符都能按顺序在t串中找到 printf("Yes\n"); else printf("No\n"); } }