Grooving Monkeys- Code Vita 8 | Pre-Qualifier Zonal Round

Grooving Monkeys

Problem Description

N monkeys are invited to a party where they start dancing. They dance in a circular formation, very similar to a Gujarati Garba
or a Drum Circle. The dance requires the monkeys to constantly change positions after every 1 second.
The change of position is not random & you, in the audience, observe a pattern. Monkeys are very disciplined & follow a specific
pattern while dancing.
Consider N = 6, and an array monkeys = {3,6,5,4,1,2}.
This array (1-indexed) is the dancing pattern. The value at monkeys[i], indicates the new of position of the monkey who is standing
at the ith position.
Given N & the array monkeys[ ], find the time after which all monkeys are in the initial positions for the 1st time.

Constraints

1<=t<=10 (test cases)
1<=N<=10000 (Number of monkeys)

Input Format

First line contains single integer t, denoting the number of test cases.
Each test case is as follows -
Integer N denoting the number of monkeys.
Next line contains N integer denoting the dancing pattern array, monkeys[].

Output

t lines,
Each line must contain a single integer T, where T is the minimum number of seconds after which all the monkeys are in their
initial position.

Test Case

Example 1
Input
1
6
3 6 5 4 1 2

Output
6

Explanation
Consider N = 6, and an array monkeys = {3,6,5,4,1,2}.
Suppose monkeys are a,b,c,d,e,f, & Initial position (at t = 0) -> a,b,c,d,e,f
At t = 1 -> e,f,a,d,c,b
a will move to 3rd position, b will move to 6th position, c will move to 5th position, d will move to 4th position, e will move to
1st position and f will move to 2nd position. Thus from a,b,c,d,e,f at t =0, we get e,f,a,d,c,b at t =1. Recursively applying same
transpositions, we get following positions for different values of t.
At t = 2 -> c,b,e,d,a,f
At t = 3 -> a,f,c,d,e,b
At t = 4 -> e,b,a,d,c,f
At t = 5 -> c,f,e,d,a,b
At t = 6 -> a,b,c,d,e,f
Since at t = 6, we got the original position, therefore the answer is 6.

Program:

#include <stdio.h>

int isequal(int x[],int m)
{
    int c1=0;
    for(int k=1;k<=m;k++)
    {
        if(x[k]==k)
        c1++;
    }
    return c1;
}
int main() {
int n,a[100],i,b[100],k,c[100],cnt,t;
scanf("%d",&t);
for(k=0;k<t;k++)
{
cnt=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[i]=i;
}
z: for(i=1;i<=n;i++)
    c[a[i]]=b[i];
cnt++;
if(isequal(c,n)==n)
printf("%d\n",cnt);
else
{
    for(i=1;i<=n;i++)
    b[i]=c[i];
    goto z;
}
}
return 0;

}


You can also run it on an online IDE: https://ide.geeksforgeeks.org/IEgUyvHl1N

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!! Cheers!!!

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