Question

我正在尝试编写一个使用k-map来解决布尔表达式的程序。我正在使用从用户接受最小项的逻辑，并检查最小项是否与我初始化的数组中的相应元素匹配，如果匹配，则使该元素1，如果不匹配，则使0。我的程序制作了所有元素0。

#include <stdio.h>
#include <conio.h>
void table(){
    printf("\n|-------------------------------------|");
    printf("\n|variables|Minterms|Term|Maxterms|Term|");
    printf("\n|---------|--------|----|--------|----|");
    printf("\n| 0| 0 | 0|x'.y'.z'| m0 |x+y+z   | M0 |");
    printf("\n| 0| 0 | 1|x'.y'.z | m1 |x+y+z'  | M1 |");
    printf("\n| 0| 1 | 0|x'.y.z' | m2 |x+y'+z  | M2 |");
    printf("\n| 0| 1 | 1|x'.y.z  | m3 |x+y'+z' | M3 |");
    printf("\n| 1| 0 | 0|x.y'.z' | m4 |x'+y+z  | M4 |");
    printf("\n| 1| 0 | 1|x.y'.z  | m5 |x'+y+z' | M5 |");
    printf("\n| 1| 1 | 0|x.y.z'  | m6 |x'+y'+z | M6 |");
    printf("\n| 1| 1 | 1|x.y.z   | m7 |x'+y'+z'| M7 |");
    printf("\n|-------------------------------------|");
}
void main()
{   int i,j,k,n,nm,ochoice,minterms[10],kmap[4][4]={0,1,3,2,
                                                4,5,7,6,
                                                12,13,15,14,
                                                8,9,11,10};
    char v[6];
    printf("\t\t\tK-MAP");
    printf("\n\t\t\t-----");
    printf("\nEnter no of variables(3or4):");
    scanf("%d",&n);
    printf("\nEnter the variables: ");
    for(i=0;i<n;i++)
    {   printf("\nEnter variable %d: ",i+1);
        scanf("%s",&v[i]);
    }
    printf("\n1.SOP");
    printf("\n2.POS");
    printf("\nEnter the required output form(1or2):-");
    scanf("%d",&ochoice);
    switch(ochoice)
    {case 1 :{system("cls");
              table();
              printf("\nEnter the number of minterms : ");
              scanf("%d",&nm);
              printf("\nEnter the corresponding minterms number from the table above : ");
              for(i=0;i<nm;i++)
              {
                  printf("\nEnter Minterm %d:",i+1);
                  scanf("%d",&minterms[i]);

              }

              for(i=0;i<4;i++)
              {
                  for(j=0;j<4;j++)
                  {
                      for(k=0;k<nm;k++){
                        if(kmap[i][j]==minterms[k]){
                            kmap[i][j]=1;
                        }
                        else{
                            kmap[i][j]=0;
                        }
                      }
                  }
              }
             for(i=0;i<4;i++){
                    printf("\n");
                for(j=0;j<4;j++){
                    printf(" %d ",kmap[i][j]);
                }
             }
             break;
             }
    }

}

Enter the number of minterms :4
Enter minterm 1:1
Enter minterm 2:5
Enter minterm 3:4
Enter minterm 4:12
output:-0 0 0 0
        0 0 0 0
        0 0 0 0
        0 0 0 0
expexted output:-0 1 0 0
                 1 1 0 0
                 1 0 0 0
                 0 0 0 0

NB：-该程序仅需一分钟即可正常工作

Answer 1

@Weather Vane指出了问题我通过制作另一个数组并对其进行更改来更正了它，这是代码

#include <stdio.h>
#include <conio.h>
void table(){
    printf("\n|-------------------------------------|");
    printf("\n|variables|Minterms|Term|Maxterms|Term|");
    printf("\n|---------|--------|----|--------|----|");
    printf("\n| 0| 0 | 0|x'.y'.z'| m0 |x+y+z   | M0 |");
    printf("\n| 0| 0 | 1|x'.y'.z | m1 |x+y+z'  | M1 |");
    printf("\n| 0| 1 | 0|x'.y.z' | m2 |x+y'+z  | M2 |");
    printf("\n| 0| 1 | 1|x'.y.z  | m3 |x+y'+z' | M3 |");
    printf("\n| 1| 0 | 0|x.y'.z' | m4 |x'+y+z  | M4 |");
    printf("\n| 1| 0 | 1|x.y'.z  | m5 |x'+y+z' | M5 |");
    printf("\n| 1| 1 | 0|x.y.z'  | m6 |x'+y'+z | M6 |");
    printf("\n| 1| 1 | 1|x.y.z   | m7 |x'+y'+z'| M7 |");
    printf("\n|-------------------------------------|");
}
void main()
{   int i,j,k,n,nm,ochoice,minterms[10],kmap[4][4]={0,1,3,2,
                                                4,5,7,6,
                                                12,13,15,14,
                                                8,9,11,10};
    int k2map[4][4]={0,0,0,0,
                     0,0,0,0,
                     0,0,0,0,
                     0,0,0,0,};

    char v[6];
    printf("\t\t\tK-MAP");
    printf("\n\t\t\t-----");
    printf("\nEnter no of variables(3or4):");
    scanf("%d",&n);
    printf("\nEnter the variables: ");
    for(i=0;i<n;i++)
    {   printf("\nEnter variable %d: ",i+1);
        scanf("%s",&v[i]);
    }
    printf("\n1.SOP");
    printf("\n2.POS");
    printf("\nEnter the required output form(1or2):-");
    scanf("%d",&ochoice);
    switch(ochoice)
    {case 1 :{system("cls");
              table();
              printf("\nEnter the number of minterms : ");
              scanf("%d",&nm);
              printf("\nEnter the corresponding minterms number from the table above : ");
              for(i=0;i<nm;i++)
              {
                  printf("\nEnter Minterm %d:",i+1);
                  scanf("%d",&minterms[i]);

              }

              for(i=0;i<4;i++)
              {
                  for(j=0;j<4;j++)
                  {
                      for(k=0;k<nm;k++){
                        if(kmap[i][j]==minterms[k]){
                            k2map[i][j]=1;

                        }
                      }
                  }
              }
             for(i=0;i<4;i++){
                    printf("\n");
                for(j=0;j<4;j++){
                    printf(" %d ",k2map[i][j]);
                }
             }
             break;
             }
    }

}

为什么该程序仅打印0？

1 个答案: