Problem Description:
In a Conference ,attendees are invited for a dinner after the conference.The Co-ordinator,Sagar arranged around round tables for dinner and want to have an impactful seating experience for the attendees.Before finalizing the seating arrangement,he wants to analyze all the possible arrangements.These are R round tables and N attendees.In case where N is an exact multiple of R,the number of attendees must be exactly N//R,,If N is not an exact multiple of R, then the distribution of attendees must be as equal as possible.Please refer to the example section before for better understanding.
For example, R = 2 and N = 3
All possible seating arrangements are
(1,2) & (3)
(1,3) & (2)
(2,3) & (1)
Attendees are numbered from 1 to N.
Input Format:
The first line contains T denoting the number of test cases.
Each test case contains two space separated integers R and N, Where R denotes the number of round tables and N denotes the number of attendees.
Output Format:
Single Integer S denoting the number of possible unique arrangements.
Constraints:
0 <= R <= 10(Integer)
0 < N <= 20 (Integer)
Sample Input 1:
1
2 5
Sample Output 1:
10
Explanation:
R = 2, N = 5
(1,2,3) & (4,5)
(1,2,4) & (3,5)
(1,2,5) & (3,4)
(1,3,4) & (2,5)
(1,3,5) & (2,4)
(1,4,5) & (2,3)
(2,3,4) & (1,5)
(2,3,5) & (1,4)
(2,4,5) & (1,3)
(3,4,5) & (1,2)
Arrangements like
(1,2,3) & (4,5)
(2,1,3) & (4,5)
(2,3,1) & (4,5) etc.
But as it is a round table, all the above arrangements are same.
Solution in C:
#include <stdio.h>
// Function to calculate factorial
long long factorial(int n) {
long long result = 1;
for (int i = 1; i <= n; i++) {
result *= i;
}
return result;
}
int main() {
int testcases;
scanf("%d", &testcases);
for (int i = 0; i < testcases; i++) {
int tables, people;
scanf("%d %d", &tables, &people);
if (tables >= people) {
printf("1\n");
} else {
int PA = people / tables;
int PB = PA + 1;
int TB = people % tables;
int TA = tables - TB;
// Using DP to store factorials pre-hand
long long fact[people + 2];
fact[0] = 1;
for (int j = 1; j <= people + 1; j++) {
fact[j] = j * fact[j - 1];
}
// Dividing people between tables
long long divide = fact[people] / (fact[PA] * fact[TB] * fact[TA] * fact[PB]);
long long result;
if (PB >= 4) {
result = divide * (factorial(PA - 1) / 2) * (factorial(TA) / 2) * (factorial(PB - 1) / 2) * (factorial(TB) / 2);
} else {
result = divide;
}
printf("%lld\n", result);
}
}
return 0;
}
You can also run it on an online IDE:
https://ide.geeksforgeeks.org/online-c-compiler/c5c2e7bb-031b-4805-9d5d-022c5f61b61c
