首页 > 代码库 > poj 1936 All in All
poj 1936 All in All
All in All
Time Limit: 1000 MS Memory Limit: 30000 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
[Submit] [Status] [Discuss]
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.
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 subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatter
Sample Output
YesNoYesNo
题意：第一个串是否是第二个的子串 区分大小写 依旧前几道题暴力的思想
#include <iostream>#include <string.h>#include <stdio.h>using namespace std;int main(){ long long int i,j; ///int会超 char s1[100000],s2[100000]; while(scanf("%s",s1)!=EOF) { scanf("%s",s2); long len1=strlen(s1); long len2=strlen(s2); i=0; j=0; while(true) { if(i==len1) { cout<<"Yes"<<endl; break; } else if(i<len1 && j==len2) { cout<<"No"<<endl; break; } if(s1[i]==s2[j]) { i++; j++; } else j++; } memset(s1,‘\0‘,sizeof(s1)); memset(s2,‘\0‘,sizeof(s2)); } return 0;}
poj 1936 All in All
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。