[结题报告]10340 - All in All Time limit: 3.000 seconds

最新推荐文章于 2025-09-10 21:59:39 发布

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最新推荐文章于 2025-09-10 21:59:39 发布

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文章标签： runtime

原文链接：http://www.cnblogs.com/sjy123/archive/2013/02/20/2919215.html

本文介绍了一个简单的算法，用于判断一个字符串是否为另一个字符串的子序列。通过实例输入输出展示了问题场景，并提供了一段C语言实现代码。

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

参考代码:
这道题给定2个字符串,求前字符串是否在后字符串出现(不要求连续),利用循环可以求出,但是这道题必须注意到的是,2个数组的定义时,务必要足够大,经证明,200000够大,定义过小,会出现Runtime error.

#include"stdio.h"
#include"string.h"
int main()
{
 char s1[200000],s2[200000];
 int i,j;
 while (scanf("%s%s",&s1,&s2)!=EOF)
 {
  for (i=0,j=0;s1[i]!='\0'&&s2[j]!='\0';j++)
  if (s1[i]==s2[j]) ++i;
  if (s1[i]=='\0') printf("Yes\n");
              else printf("No\n");
 }
 return 0;
}