poj 3302 Subsequence

最新推荐文章于 2020-10-31 08:45:27 发布

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最新推荐文章于 2020-10-31 08:45:27 发布

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/23/2259805.html

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6865		Accepted: 3998

Description

Given a string s of length n, a subsequence of it, is defined as another string s' = s_u_₁s_u_₂...s_{u_m} where 1 ≤ u₁ < u₂ < ... < u_m ≤ n and s_i is the ith character of s. Your task is to write a program that, given two strings s1 and s2, checks whether either s2 or its reverse is a subsequence of s1 or not.

Input

The first line of input contains an integer T, which is the number of test cases. Each of the next T lines contains two non-empty strings s1 and s2 (with length at most 100) consisted of only alpha-numeric characters and separated from each other by a single space.

Output

For each test case, your program must output "YES", in a single line, if either s2 or its reverse is a subsequence of s1. Otherwise your program should write "NO".

Sample Input

5
arash aah
arash hsr
kick kkc
A a
a12340b b31

Sample Output

YES
YES
NO
NO
YES

#include<iostream>
using namespace std;
int main()
{
int n;
char a[101],b[101];
int i,j;
int lena,lenb;
int mark;
int res1,res2;
    cin>>n;
while(n--)
    {
        cin>>a;
        cin>>b;
        lena=strlen(a);
        lenb=strlen(b);
        mark=0;
for(j=0;j<lenb;j++)
        {
for(i=mark;i<lena;i++)
            {
if(b[j]==a[i])
break;
            }
if(i==lena)
            {
                res1=0;
break;
            }
else
                mark=i+1;
        }
if(j==lenb)
            res1=1;
        mark=0;
for(j=lenb-1;j>=0;j--)
        {
for(i=mark;i<lena;i++)
            {
if(b[j]==a[i])
break;
            }
if(i==lena)
            {
                res2=0;
break;
            }
else
                mark=i+1;
        }
if(j==-1)
            res2=1;
if(res1==0&&res2==0)
            cout<<"NO"<<endl;
else
            cout<<"YES"<<endl;
    }
return 0;
}