POJ 1936 All in All

简介: DescriptionYou have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way.

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
题目要求就是在s2中找字串s1!

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
char a[100005],b[100005];
int main()
{
    while(~scanf("%s",a)){
        scanf("%s",b);
        int a_length = strlen(a);
        int b_length = strlen(b);
        int a1,b1=0;
        int i;
        for(i=0;i<a_length;i++){
                bool flag=false;
            for(int j=b1;j<b_length;j++){
                if(a[i]==b[j]){
                    b1=j+1;
                    flag=true;
                    break;
                }
            }
            if(!flag)
                break;
        }
        if(i==a_length)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}
目录
相关文章
|
9月前
|
算法
Highways(POJ—2485)
Highways(POJ—2485)
|
9月前
|
算法 数据建模
Poj 3169(差分约束系统)
Poj 3169(差分约束系统)
39 0
POJ 1804
题目:http://poj.org/problem?id=1804 大意:给你一串数字,排序。求出最少的交换次数  \ 我用归并做的 #include #include using namespace std; int aa[500010],bb[500010]; long lon...
710 0
|
人工智能 BI
poj-1008-玛雅历
Description 上周末,M.A. Ya教授对古老的玛雅有了一个重大发现。从一个古老的节绳(玛雅人用于记事的工具)中,教授发现玛雅人使用了一个一年有365天的叫做Haab的历法。这个Haab历法拥有19个月,在开始的18个月,一个月有20天,月份的名字分别是pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu。
893 0
|
机器学习/深度学习 算法
|
机器学习/深度学习
|
机器学习/深度学习