#include <stdio.h>
 #define MIN 2
 #define MAX 5
 int pRbar ( double area, int *size, int *no );

 int main() {
     int s, n, i;
     double area, cc;

     area = 15.00;
     while ( area <= 105.00) {
         i = pRbar ( area, &s, &n );
         if ( i == 1 )
             printf("To get %.0lf, use %d-#%d\n", area, n, s );
         else
             printf("No match found\n");
         area += 15.00;
     }
     return 0;
 }
 int pRbar ( double area, int *size, int *no ) {
     int i, j, k;
     double a;

     k = 1;
     for ( j = MIN; j <= MAX && k; j++ ) //j <= MAX && k不就是j<=5&&1，这不是自相矛盾吗？
     {
          for ( i = 1; i <= 4 && k; i++ )//疑问和上一个一样
            {
               switch ( i ) {
               case 1:
                   a = 7.50;
                   break;
               case 2:
                   a = 11.50;
                   break;
               case 3:
                   a = 15.25;
                   break;
               case 4:
                   a = 19.05;
                   break;
               }
               if (a * j >= area) {
                   k = 0;
                   *size = i;
                   *no = j;
               }
          }
     }
     printf("(%d-#%d %5.2lf)\n", j-1, i-1, a);
     return !k;//除了k之外其他都输出吗？
 }

for ( j = MIN; j <= MAX && k; j++ ) //j <= MAX && k不就是j<=5&&1，这不是自相矛盾吗？
return !k;//除了k之外其他都输出吗？
我的疑问是这两个。

我问的重点不是k的值，是怎么理解j <= MAX && k，之前k=1，max=5