Beautiful 3 Set


Problem Statement :


You are given an integer n. A set, S, of triples (xi,yi,zi) is beautiful if and only if:
0 <= xi,yi,zi
xi+yi+zi = n, for all i : 1 <= i <= |S|
Let X be the set of different xi's in S, Y be the set of different yi's in S, and Z be the set of different zi in S. Then |X| = |Y| = |Z| = |S|.
The third condition means that all xi's are pairwise distinct. The same goes for yi and zi.

Given , find any beautiful set having a maximum number of elements. Then print the cardinality of  (i.e., ) on a new line, followed by  lines where each line contains  space-separated integers describing the respective values of , , and .

Input Format

A single integer, n.

Constraints
1 <= n <= 300
Output Format

On the first line, print the cardinality of S (i.e., |S|).
For each of the |S| subsequent lines, print three space-separated numbers per line describing the respective values of xi, yi, and zi for triple i in S.


Solution :



title-img


                            Solution in C :

In C++ :





#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int N;
    cin >> N;
    if (N == 1){
        cout << "1\n0 0 1\n";
    } else if (N == 2){
        cout << "2\n0 0 2\n1 1 0\n";
    } else if (N == 3){
        cout << "3\n0 1 2\n2 0 1\n1 2 0\n";
    } else if (N == 4){
        cout << "3\n0 0 4\n1 1 2\n2 2 0\n";
    } else {
        if (N % 3 == 0){
            int h = N/3*2+1;
            cout << h << "\n";
            for (int i=0; i<h; i++){
                cout << i << ' ' << (N/3+i)%h << ' ' << (N - i - (N/3+i)%h) << '\n';
            }
        } else if (N % 3 == 2){
            int h = N-N/3;
            cout << h << "\n0 0 " << N << "\n";
            for (int i=1; i<h; i++){
                int m = i <= h/2 ? (i + N/3) : (i - N/3 - 1);
                cout << i << ' ' << m << ' ' << (N-i-m) << "\n";
            }
        } else {
            int h = N-N/3;
            cout << h << "\n0 0 " << N << "\n1 1 " << N-2 << "\n";
            for (int i=1; i<=N/3; i++){
                cout << i+1 << ' ' << i+N/3 << ' ' << N - (i+1) - (i+N/3) << '\n';
            }
            for (int i=1; i<N/3; i++){
                cout << i+N/3+1 << ' ' << i+1 << ' ' << N - (i+N/3+1) - (i+1) << '\n';
            }        
        }
    }
    return 0;
}









In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int max = n/3*2;
        if (n%3==2)
            max++;
        int sn = n;
        if (n%3==1)
            sn--;
        System.out.println(max+1);
        int first = max;
        for (int i = (max+1)/2; i >= 0; i--) {
            System.out.println(first+" "+i+" "+(n-i-first));
            first--;
        }
        for (int i = sn-first-1; first >= 0; i--) {
            System.out.println(first+" "+i+" "+(n-i-first));
            first--;
        }
    }
}









In C :






#include <stdio.h>
#include <stdlib.h>



int main() {

    int n, k, r, **S, i, q;

    scanf("%d",&n);

	k = (2*n)/3;
	r = (2*n)%3;

	S = malloc(sizeof(int*)*k);

	switch ( r ) {

		case 0:

			for (i=0; i<=k/2; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i;
				S[i][1] = k-2*i;
				S[i][2] = k/2+i;

			}

			for (i=k/2+1; i<=k; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i;
				S[i][1] = 2*(k-i)+1;
				S[i][2] = i-k/2-1;

			}

			break;

		case 2:

			for (i=0; i<=k/2; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i;
				S[i][1] = k-2*i;
				S[i][2] = k/2+i+1;

			}

			for (i=k/2+1; i<=k; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i;
				S[i][1] = 2*(k-i)+1;
				S[i][2] = i-k/2;

			}

			break;

		case 1:

			q = k/2;

			for (i=0; i<=q; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i;
				S[i][1] = k-2*i;
				S[i][2] = q+i+1;

			}

			for (i=q+1; i<=k; i++) {

				S[i] = malloc(sizeof(int)*3);

				S[i][0] = i+1;
				S[i][1] = 2*(k-i);
				S[i][2] = i-q-1;

			}

			break;

		default:
			break;

	}

    printf("%d\n",k+1);
    for (i=0; i<=k; i++)
        printf("%d %d %d\n", S[i][0], S[i][1], S[i][2]);

    return 0;

}









In Python3 :





n = int(input().strip())

low = n // 3
high = 2 * n // 3

print(high + 1)

for i in range(low+1):
    print(i, 2*(low-i), n+i-2*low)
    
for j in range(low+1, high+1):
    print(j, 2*(high-j)+1, n+j-1-2*high)
                        




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