poj 3913（水）

时间：2016-08-26 10:22:25 阅读：217 评论：0 收藏：0 [点我收藏+]

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
题意：递增或递减

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 typedef long long ll;
 6 #include<cstring>
 7 #define N 200
 8 using namespace std;
 9 int main()
10 {
11     int t;
12     scanf("%d",&t);
13     int a[N],b[N],c[N];
14     for(int i=0;i<t;i++)
15     {
16         scanf("%d%d%d",&a[i],&b[i],&c[i]);
17     }
18     printf("Gnomes:\n");
19     for(int j=0;j<t;j++)
20     {
21         if((a[j]<b[j]&&b[j]<c[j])||(a[j]>b[j]&&b[j]>c[j]))
22             printf("Ordered\n");
23         else
24             printf("Unordered\n");
25     }
26 }

poj 3913（水）

原文：http://www.cnblogs.com/Aa1039510121/p/5809199.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年09月23日 (328)
2021年09月24日 (313)
2021年09月17日 (191)
2021年09月15日 (369)
2021年09月16日 (411)
2021年09月13日 (439)
2021年09月11日 (398)
2021年09月12日 (393)
2021年09月10日 (160)
2021年09月08日 (222)