Battle plague Cup -- strange shape

Xiao Fan has one with a side length of 2 Square white paper , It's painted 2×2 The length of each side is 1 Lattice of , Each grid can be placed with an edge length of 1 The cube . There are several edges in Xiao Fan's hand, the length of which is 1 The cube , He randomly placed some cubes on each grid , Find the surface area of the geometry composed of these cubes .

Input format :

Enter two lines , Two per line, no more than 100 The positive integer , Stands for looking down at , The number of cubes on each grid drawn on white paper .

Output format :

Output one line , Contains an integer , Represents the surface area of the geometry .

sample input :

The input data corresponding to the above figure is as follows ：

2 1
1 1

sample output :

20

The code is as follows ：

#include<iostream>
using namespace std;
int main() {
	int a[5];
	cin>>a[1]>>a[2]>>a[3]>>a[4];
	int sum = 0;
	sum = (a[1]+a[2]+a[3]+a[4])*6 - 8;// Because it is a positive integer , Therefore, there are at least 4 Square , Each adjacent one is reduced by two faces ,4 Two adjacent , So reduce 8 Face to face , So subtract 8
	for (int i = 1; i < 5; i++) {
		if (a[i] > 1) sum -= 2*(a[i]-1);
	}// Every additional layer , Contact with the bottom layer , So reduce 2 Face to face 
	int max = (a[1]>a[2])?a[1]:a[2];// If input 4 3 1 2, Then the top layer is not equal , But the middle layer has equal , So we should judge by circulation 
	int min = (a[1]<a[2])?a[1]:a[2];
	for (;max > 1; max--) {
		if (max == min) {
			sum -= 2 * (max - 1);
		}
	}// In the first row, two squares are adjacent 
    max = (a[1]>a[3])?a[1]:a[3];
    min = (a[1]<a[3])?a[1]:a[3];
    for (;max > 1;max--) {
		if (max == min) sum -= 2 * (max - 1);
	}// In the first column, two squares are adjacent 
	max = (a[3]>a[4])?a[3]:a[4];
    min = (a[3]<a[4])?a[3]:a[4];
	for (;max > 1;max--) {
		if (max == min) sum -= 2 * (max - 1);
		}// In the second row, two squares are adjacent 
    max = (a[2]>a[4])?a[2]:a[4];
    min = (a[2]<a[4])?a[2]:a[4];
	for (;max > 1;max--) {
     	if (max == min) sum -= 2 * (max - 1);
		}// In the second column, two squares are adjacent 
	cout<<sum;
}
 min) sum -= 2 * (max - 1);
		}// In the second column, two squares are adjacent 
	cout<<sum;
}