Realization of truth table assignment by discrete mathematical programming

Code ：

#include<stdio.h>
#include<math.h>
#define LEN 666 // Define array length 
int arr[LEN];   // For storage 2^n Sub characters ,n Represents the number of variables 
int brr[LEN][4+10];   //brr Used to store truth tables 
int beg = 80;             // character P Corresponding ASCII code 
int sta1 = 0, sta2 = 0;   //sta1 Indicates that the true value is 1 The number of ,sta2 Indicates that the true value is 0 The number of 

int main()
{
    
    int num;  // Number of variables 
    printf(" Please enter the number of variables  : ");
    scanf("%d",&num);
    int sum = pow(2, num);  //2^n
	printf(" Please enter  %d  A truth value ( With (p->q)->r For example ) : ",sum); // Input (p->q)->r The true value of , Examples can be changed freely 
    for(int i = 1; i <= sum; i++)
    {
    
        scanf("%d",&arr[i]);
        if(arr[i] == 1)
            sta1++;  //sta1 Indicates that the true value is T The number of 
        else
            sta2++;  //sta2 Indicates that the true value is F The number of 
    }
    printf("\n");
    // Processing truth table 
    int cnt1 = 0, cnt2 = 1; 
    for(int i = sum-1; i >= 0; i--)  // Binary implementation of the truth table assignment 
    {
    
        cnt1 = 0;
        int val=i;
        while(cnt1 < num)
        {
    
            cnt1++;
            brr[cnt2][cnt1] = val%2;
            val = val/2;
        }
        cnt2++;
    }
	printf(" Output the truth table corresponding to the formula  :\n");
    for(int i = 1; i <= num ;i++)
    {
    
		printf("%c\t",beg++);
    }
    printf("(p->q)->r\n");
    beg = 80;   
    for(int i = 1; i <= sum; i++)
    {
    
        for(int j = num; j >= 1; j--)
        {
    
            printf("%d\t",brr[i][j]);
        }
        printf("%d\n",arr[i]);
    }
    printf("\n");
    int k = 0;

	printf(" Output main disjunctive normal form :\n");
    for(int i = 1; i <= sum; i++)
    {
    
        if(arr[i] == 1)
        {
    
            k++;
            printf(" ( ");
            for(int j = num; j >= 1; j--)
            {
    
                if(brr[i][j] == 1)
					printf("%c",beg++);
                else
					printf("┓ %c",beg++);
                if(j != 1)
				printf(" ∧ ");
            }
            printf(" ) ");
            if(k < sta1)
				printf(" ∨ ");
        }
        beg = 80;
    }
    printf("\n\n");
         
	k=0;
    printf(" Output principal conjunctive normal form :\n");
    for(int i = 1; i <= sum; i++)
    {
    
        if(arr[i] == 0)
        {
    
            k++;
            printf(" ( ");
            for(int j = num; j >= 1; j--)
            {
    
                if(brr[i][j] == 0)
                {
    
                    printf("%c",beg++);
                }
                else
                {
    
                    printf("┓ %c",beg++);
                }
                if(j != 1)
                    printf(" ∨ ");
            }
            printf(" ) ");
            if(k < sta2)
                printf(" ∧ ");
        }
        beg = 80;
    }
    printf("\n\n");

    return 0;
}

Examples

Running results ：