Two full permutation enumeration methods

Topic description

Two solutions are provided below, both of which are essentially all permutations of enumeration

Solution one

#include <iostream>
#include <algorithm>
using namespace std;
int ans;
bool st[10];
int p[10];

void dfs(int u) {
    
    
	if (u > 8) {
    
    
		int x = p[0] + p[1] + p[2] + p[3];
		int y = p[3] + p[4] + p[5] + p[6];
		int z = p[6] + p[7] + p[8] + p[0];
		if (x == y && y == z ) ans ++;
		return;
	}
	
	for (int i = 1; i <= 9; i++) {
    
    
		if (!st[i]) {
    
    
			st[i] = true;
			p[u] = i;
			dfs(u+1);
			st[i] = false;
		}
	}
}

int main() {
    
    
	dfs(0);
	cout << ans / 6; // 144
	
	return 0;
}

Solution two

#include <iostream>
#include <algorithm>
using namespace std;
int ans;

int main() {
    
    
	int a[9] = {
    
    1, 2, 3, 4, 5, 6, 7, 8, 9};
	
	int x = a[0] + a[1] + a[2] + a[3];
	int y = a[3] + a[4] + a[5] + a[6];
	int z = a[6] + a[7] + a[8] + a[0];
	
	if (x == y && y == z ) ans ++;
	
	while (next_permutation(a, a+9)) {
    
    
		int x = a[0] + a[1] + a[2] + a[3];
		int y = a[3] + a[4] + a[5] + a[6];
		int z = a[6] + a[7] + a[8] + a[0];
	
		if (x == y && y == z ) ans ++;
	}
	cout << ans / 6;
	
	return 0;
}