UVA10340 All in All子序列

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


输入两个字符串s和t，判断是否可以从t中删除一个或多个字符（其他字符顺序不变），得到s。

 1 #include<iostream>
 2 
 3 int main()
 4 {
 5     using namespace std;
 6     char a[100010], b[100010];
 7     int num1 = strlen(a);
 8     int num2 = strlen(b);
 9     int i, j;
10     cin >> a >> b;
11     for (i = 0, j = 0; i < num1&&j < num2;)
12     {
13         
14         if (a[i] == b[j])     //满足条件  i加1，j加1
15         {
16             i++; j++;
17         }
18         else j++;       //否则 j+1，直到找到 a[i]==b[j]
19         
20     }
21     if (i == num1)
22         cout << "bingo!";
23     else cout << "wrong!";
24     system("pause");
25         return 0;
26 }

UVA10340 All in All子序列

原文：http://www.cnblogs.com/ghost-song/p/4458586.html