11th C++ problem

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Come on Guys this is not that much tough as compared to 10th
or you guys are not interested to solve these

goyal420Example
Input:
32
0 0
3
3 2 1
6
4 8 6 1 5 2
Output:
1
2
4
Output details
Case 1: No matter where who reads there number first, just one person is eligible for the
lucky draw.
Case 2: The person in the last position could read their number first to obtain the sequence
1 3 2. We could then pick 2 sheets with numbers 1 and 2.
Case 3: The person in the fourth position could read their number first to obtain the
sequence 1 5 2 4 8 6. We could then pick 4 sheets with numbers 1 2 4 6

The sample input seems to be wrong.
The first line should be 2, right? It is given as 32.

Also, please explain the output details. I couldn't get what you mean by the cases 1,2 and 3.

Thanks for giving attention ss
yes you are right it is
3
2
00
..
In all the cases the person with smallest no will read it's no first and then the next person in clock wise direction

so here The First case implies since there are two persons hence it is quite obvious that only one person is eligible for lucky draw
2 case:-here is it given 3 2 1 hence person at 3rd position will read its no first
and the sequence becomes 1 3 2 Now among these only those persons are valid which forms increasing sequence ie 1 2 .hence only two persons are valid for luck draw

3:-case same for this 4 8 6 1 5 2
the sequence will be 1 5 2 4 8 6
thus here persons for draw are 1 2 4 6 ie 4 hence output should be 4

I get it now. I'll try to program it today..

so here is the solution for this

#include<iostream.h>
#include<conio.h>
int getvalue(int a[],int b[],int );
int small(int a[],int);
void main()
{
clrscr();
int n,i,j,count=1;
int *store;
cin>>n;
store=new int[n];
int c[100],d[100];
for(i=0;i<n;i++)
{
int b=0,*c,*d;
cin>>b;
c=new int[b];
d=new int[b];
store[i]=getvalue(c,d,b);
delete []c;
delete []d;
}
for(i=0;i<n;i++)
{
cout<<store[i]<<endl;
}
delete []store;
getch();
}
int getvalue(int c[],int d[],int b)
{
int j,i,sm,count=1;
for(j=0;j<b;j++)
{
cin>>c[j];
}
sm=small(c,b);
for(i=0;i<b;i++)
    {
d[i]=c[sm];
if(sm==b-1)
{
sm=0;
}
else
{
sm++;
}
}
for(i=b-1;i>0;i--)
    {
        if(d[i]>d[i-1])
        {
        count++;
        }
    }

for(i=0;i<b;i++)
{
c[i]=0;
d[i]=0;
}
return count;
}
int small(int a[],int n)
    {
    int s = a[0];
    int pos, i;
    for (i = 0; i < n; i++)
    {
        if (a[i] <= s)
        {
            s = a[i];
            pos = i;
        }
    }
    
    return pos;
}