POJ 1936(字符串匹配) 解题报告

/*_______________________________________________POJ 1936题___________________________________________________________ All in All Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 18702 Accepted: 7442 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 __________________________________________________________________________________________________________________*/ #include #include int main() { int i,j,m,n; char s[100000]={'/0'}, t[100000]={'/0'}; // FILE *fin=fopen("input.txt","r"); while( scanf("%s%s",s,t)==2 ) { m=strlen(s); //注：一定要提前计算好长度，不要每次循环判断条件中都临时计算长度，太费时间 n=strlen(t); if(m>n) { printf("No/n"); continue; } for(i=0,j=0; i