PROBLEM LINK:

Practice
Contest

Author:Konstantin Sokol
Tester:Tasnim Imran Sunny
Editorialist:Praveen Dhinwa

DIFFICULTY:

CAKEWALK

PREREQUISITES:

simple combinatorics.

PROBLEM:

Given following 5 patterns of a flag.

112233
112233
112233

111111
222222
333333

112222
112222
113333

122223
111333
144443

111222
333222
333444

Your need to count the number of different well-painted flags. We call a flag well-painted if it's made according to the following algorithm:

• Pick up one of the flag patterns considered above;
• Paint each connected component in an color encoded by an integer from 1 to N. Different colors are encoded with different integers. If two different connected components share a common side(not corner), than they must be painted in different colors. In any other case they can be painted in both equal and different colors.

QUICK EXPLANATION

• For each flag, we can find out the number of different well painted flags. The flags are given in such a way that no two flags in the 5 flags will be similar to each other. So you can solve the problem independently for each flag.
• For a single flag, find number of ways of painting that flag using N colors.

EXPLANATION

For each flag, we can find out the number of different well painted flags independently because the shape of the flags are in such a way that no two flags in the above given 5 flags will be similar to each other in any possible painting. So you can solve the problem independently for each flag.

Flag 1
112233
112233
112233

We can use minimum 2 colors to make the flag well painted.
112211
112211
112211

So here we can paint the first connected component by using any of the N possible colors. Then for second connected component, total possible colors available will be N - 1 because we can not use the color of first connected component. Then for third connected component, total possible colors available will also be N - 1 because we can not use color of second connected component.

Hence number of ways = N * (N - 1) * (N - 1).

Flag 2
111111
222222
333333

We can use minimum 2 colors to make the flag well painted.
111111
222222
111111

So here we can paint the first connected component by using any of the N possible colors. Then for second connected component, total possible colors available will be N - 1 because we can not use the color of first connected component. Then for third connected component, total possible colors available will also be N - 1 because we can not use color of second connected component.

Hence number of ways = N * (N - 1) * (N - 1).

Flag 3
112222
112222
113333

We can use minimum 3 colors to make the flag well painted.
112222
112222
113333

So here we can paint the first connected component by using any of the N possible colors. Then for second connected component, total possible colors available will be N - 1 because we can not use the color of first connected component. Then for third connected component, total possible colors available will also be N - 2 because we can not use color of first and second connected component.

Hence number of ways = N * (N - 1) * (N - 2).

Flag 4
122223
111333
144443

We can use minimum 3 colors to make the flag well painted.
122223
111333
122223

So here we can paint the first connected component by using any of the N possible colors. Then for second connected component, total possible colors available will be N - 1 because we can not use the color of first connected component. Then for third connected component, total possible colors available will also be N - 2 because we can not use color of first and second connected component. Then for the fourth connected component, total numbers of colors available will be N - 2 because we can not use first and thrid connected component.

Hence number of ways = N * (N - 1) * (N - 2) * (N - 2).

Flag 5
111222
333222
333444

We can use minimum 3 colors to make the flag well painted.
111222
333222
333111

So here we can paint the first connected component by using any of the N possible colors. Then for second connected component, total possible colors available will be N - 1 because we can not use the color of first connected component. Then for third connected component, total possible colors available will also be N - 2 because we can not use color of first and second connected component. Then for the fourth connected component, total numbers of colors available will be N - 2 because we can not use second and thrid connected component.

Hence number of ways = N * (N - 1) * (N - 2) * (N - 2).

Final answer

• For flag 1: N * (N - 1) * (N - 1).
• For flag 2: N * (N - 1) * (N - 1).
• For flag 3: N * (N - 1) * (N - 2).
• For flag 4: N * (N - 1) * (N - 2) * (N - 2)
• For flag 5: N * (N - 1) * (N - 2) * (N - 2)

Pseudo code

long long n;
read n
long long ans = 2*n*(n-1)*(n-1) + n*(n-1)*(n-2) + 2*n*(n-1)*(n-2)*(n-2);
print ans

Complexity
We just need constant amount of addition and multiplication operations. Hence the time complexity will be O(1).

AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution
Tester's solution