All in All（模拟）

C - All in AllTime Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d& %I64uSubmit Status Practice POJ1936Appoint description: System Crawler  (2015-10-03)DescriptionYou 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 intothe 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. InputThe 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.OutputFor 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

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<vector>#include<cmath>using namespace std;#define T 100005#define inf 0x3f3f3f3f#define CRL(a) memset(a,inf,sizeof(a))typedef long long ll;int main(){/*freopen("input.txt","r",stdin);*/char s[T],t[T];int i,j,k;while(~scanf("%s%s",&s,&t)){for(i=0,k=0;t[i]&&s[k];++i){if(s[k]==t[i]){k++;}}if(s[k]=='\0')printf("Yes\n");elseprintf("No\n");}return 0;}