POJ 1936(字符串匹配) 解题报告

/*_______________________________________________POJ 1936题___________________________________________________________All in AllTime 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 Inputsequence subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatterSample OutputYesNoYesNo__________________________________________________________________________________________________________________*/#include<stdio.h>#include<string.h>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<m && j<n ;){if(s[i]==t[j]){i++;j++;}else{j++;}}if(j=n && i<m ) //t串扫描完而s串还没扫完printf("No/n");elseprintf("Yes/n");}return 0;}//(1)测试例 “aa a”提醒我应该先比较s t 的长度 //(2)第一版因为将strlen(s)写到了循环判断条件里，导致超时//(3)竟然将Yes No写成YES NO