Help..getting WA

general
,
1

In which testcases am i getting WA?
link text

here’s my solution:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main()
{
   ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    long t;
    cin>>t;
    while(t--)
    {
       long a[10001],b[10001],m,n,flag=0,i,j;
       cin>>m;
       for(i=1;i<=m;++i) cin>>a[i];
       	cin>>n;
       for(i=1;i<=n;++i) cin>>b[i];
       	for(i=1;i<=m;++i)
       	{
       		int count=0,k=1;
       		if(a[i]==b[k])
       		{
       			k=i;++count;
       			for(j=2;j<=n;++j)
       			{
                   if(a[++k]==b[j]) ++count;
                   else break;
       			}
       			if(count>=n) {cout<<"Yes\n";flag=1;break;}
       		}
       	}
       	if(flag==0) cout<<"No\n";
    }
    return 0;
}
2

Please write a more systematic solution. It’s very hard to even read this solution of yours.

3

Bro edit and Format code… its very difficult for understanding and help

4

ORDER DOES NOT MATTER IN THAT Q

Just check if elements of shorter sequence are present in longer one. That Q is ambiguous.

In other words, DONT look for order!! Just check if content of fav sequence are present or not. Should give you AC.

5

yeah did that…still getting WA

6

K, wait, will debug ur code in 10 min.

7

got AC thankyou…

8

MEanwhile, heres my AC code if it helps you https://www.codechef.com/viewsolution/13785476 (ignore the “check()” function i made. It has no relevance to the code)

9

thank you!!

10

Your code is order sensitive. For example, in-

6
1 2 3 4 5 6
3
2 3 4

It gives yes. It should also give yes for ‘4 3 2’ then. But its giving a no. What you can do, is, reset the loop variable to start of array. Once you find at a[i]==b[j] , you can change the a[i] to some value like -1 so it doesnt get counted more than once. This should help ^^