### PROBLEM LINK:

Practice

Contest

**Author:** Misha Chorniy

**Tester:** Karan Aggarwal

**Editorialist:** Pushkar Mishra

### DIFFICULTY:

Simple

### PREREQUISITES:

None

### PROBLEM:

Given four numbers a, b, c, d, tell whether it is possible to pair them up such that a:b is equal to c:d. We are allowed to shuffle the order of the numbers.

### EXPLANATION:

We can simply try all the possible pairings. A way to model this is to cycle through all the permutations of the four numbers, pair up the first two together and the last two together. Then find the ratio of the first two and the last two; if they are equal, output â€śPossibleâ€ť, else â€śImpossibleâ€ť. Cycling through permutations can be done through functions like next\_permutation in the C library or simply by recursion. Either way works since we just have 4 numbers.

A more intelligent solution is to sort the four numbers and pair up the first 2 together and the last 2 together and check their ratios. This works because ratios are symmetric.

For checking the ratio equality, we can use an simple property that if a:b = c:d then a*d = b*c. This way, we can avoid dealing with floats.

Please see editorialistâ€™s/setterâ€™s program for implementation details.

### COMPLEXITY:

\mathcal{O}(1)

### SAMPLE SOLUTIONS:

Author

Tester

Editorialist

Admin

2 Likes

â€śCycling through permutations can be done through functions like next_permutation in the C library or simply by recursion.â€ť

Or with 4 copypasted cycles + checking if they give a permutation (which can be done with just 4 conditions - three indices must be different and the sum of all 4 must be 0+1+2+3). In this case, it should be the simplest solution.

2 Likes

can anyone plz tell the case this code fails

```
#include<iostream>
using namespace std;
#include<algorithm>
#include<vector>
main()
{
vector <int> v;
for(int i=0;i<4;i++)
{
int temp;
cin>>temp;
v.push_back(temp);
}
sort(v.begin(),v.end());
int k1=v[1]/v[0];
int k2=v[3]/v[2];
int r1=v[1]%v[0];
int r2=v[3]%v[2];
if(k1==k2 && r1==r2)
{
cout << "Possible";
}
else
{
cout<<"Impossible";
}
}
```

does these both

a/b == c/d and a:b == c:d

are equal in case of this question ??

because i was getting WA with a/b == c/d.

Integer division. 10/4 and 10/5 are equal according to your submission.

saikarthik12

Try this: 3 5 6 8 the answer should be impossible

btw my answer was just an if condition the problem isnâ€™t that big of a deal

1 Like

It will fail when a/b is a floating number.

eg - 1 2 1 4.

Answer should be impossible but your code will give possible as your ratios are 0 and 0 whereas they should be 0.2 and 0.25.

#include

using namespace std;

int main()

{

int a,b,c,d;

cin>>a>>b>>c>>d;

if((a*b==c*d) || (a*c==b*d) || (a*d==b*c))

{

cout<<â€śPossibleâ€ť<<endl;

```
}
else
cout<<"impossible";
```

}

what is wrong in my code?

@alok12345

â€śImpossibleâ€ť not â€śimpossibleâ€ť. Case is important though I assume this may already have been solved by you xD.