All in All（送）

最新推荐文章于 2018-09-08 13:20:50 发布

原创最新推荐文章于 2018-09-08 13:20:50 发布 · 396 阅读

CC 4.0 BY-SA版权

本文探讨了一种新颖的加密方法，通过在消息中插入随机生成的字符串来编码信息。虽然具体的生成和插入机制未详细阐述，但提供了一个程序来验证最终字符串中是否确实包含了原始消息的编码版本。详细解释了如何通过比较两个字符串来判断一个是否为另一个的子序列，进而验证加密的有效性。

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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. Input is terminated by EOF.
Output
For each test case output, if s is a subsequence of t.
Sample Input
sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter
Sample Output
Yes
No
Yes
No
题意：第二个串中的字符是否在第一个串里按顺序出现，就是子串的弱化版。
O（n）算法，直接扫一遍。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
char a[1000005],b[1000005];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    while(scanf("%s%s",a,b) != EOF)
    {
        int i,j,len1,len2;
        i = j = 0;
        len1 = strlen(a);len2 = strlen(b);
        while(i < len1&&j < len2)
        {
            if(a[i] == b[j])
            {
                i++;j++;
            }
            else
            {
                j++;
            }
        }
        if(i == len1)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}