UVA - 10340 All in All

最新推荐文章于 2021-03-20 10:18:45 发布

acm_1361677193

最新推荐文章于 2021-03-20 10:18:45 发布

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本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/41146915

本文介绍了一个简单的算法问题，即判断一个字符串是否为另一个字符串的子序列。通过两个示例代码展示了如何实现这一功能，适用于编程竞赛及算法验证。

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All in All

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem E

All in All

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

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 Specification

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 Specification

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

Source: ULM Local Contest

分析：

刘汝佳大白书里的水题。一开始居然把Yes和No写错了，结果错了几发，仔细看了一下，改了过来，就AC了。

uva又挂了，在poj1936交了一发（跟uva上的题一样，并且更详细，比如，规定：The length of s and t will no more than 100000.），ac了。

ac代码：（1）

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100005;
char s[maxn],t[maxn];
int main()
{
  while(scanf("%s%s",s,t)!=EOF)
  {
      int n=strlen(s),m=strlen(t);
      /*if(n>m)
      {
          printf("No\n");
          continue;
      }*/
      int i;
      int j=0;
      for(i=0;i<n;i++)
      {
          for(;j<m;j++)
          {
              if(s[i]==t[j])
              {
                  j++;break;
              }
          }
          if(j==m) break;
      }
      if(i==n||i==n-1&&s[n-1]==t[m-1]) printf("Yes\n");
      else printf("No\n");
  }
    return 0;
}

ac代码：（2）

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100005;
char s[maxn],t[maxn];
int main()
{
  while(scanf("%s%s",s,t)!=EOF)
  {
      int n=strlen(s),m=strlen(t);
      if(n>m)
      {
          printf("No\n");
          continue;
      }
      int i;
      int j=0;
      for(i=0;i<m;i++)
      {
         if(t[i]==s[j]&&j++==n-1)
         {
             break;
         }
      }
      if(j==n) printf("Yes\n");
      else printf("No\n");
  }
    return 0;
}