我们可以用分治的方法来解决这个问题。
想象我们的输入= [a，B，c，d]
我们想从[a，B]和[c，d]中找到答案
然后合并最后的解决方案以找到答案。

if(a > b) swap(&a,&b); // solve [a,b]
if(c > d) swap(&c,&d); // solve [c,d]
if(a > c) swap(&a,&c); // find minimum from [a,b] and [c,d]
if(d > b) swap(&b,&d); // find maximum from [a,b] and [c,d]

// a will store the minimum value.
// b will store the maximum value.

附加（如何在C语言中交换数字）

void swap(int *a,int *b) {
    int c = *a;
    *a = *b;
    *b = c;
}

下面应该工作。
它使用前2个if来计算max1 = max(a, b)和max2 = max(c, d)，以及min1 = min(a, b)和min2 = min(c, d)。
则最大值等于max(max(a, b), max(c, d)) = max(max1, max2)（第三个if）。
最小值等于min(min(a, b), min(c, d)) = min(min1, min2)（第四个if）。

int a, b, c, d;

// ...

int min, max, max1, max2, min1, min2;

if (a > b) // 1
{
    max1 = a;
    min1 = b;
}
else
{
    max1 = b;
    min1 = a;
}

if (c > d) // 2
{
    max2 = c;
    min2 = d;
}
else
{
    max2 = d;
    min2 = c;
}

if (max1 > max2) // 3
    max = max1;
else
    max = max2;

if (min1 < min2) // 4
    min = min1;
else
    min = min2;

奖金

零if秒。

int a, b, c, d;

// ...

int min = a, max = a;
b < min && (min = b);
c < min && (min = c);
d < min && (min = d);
b > max && (max = b);
c > max && (max = c);
d > max && (max = d);

#include <stdio.h>

int main ()
{
    int num1, num2, num3, num4;
  
     printf("enter the numbers: ");
     scanf("%d %d %d %d", &num1, &num2, &num3, &num4);

     int large1 = num1, low1 = num2, low = num3, large = num4;
     if (num1 <= num2) {
         large1 = num2;
         low1 = num1;
     }
     if (num3 >= num4) {
         large = num3;
         low = num4;
     }
     if (large < large) {
         large = large1;
     }
     if (low > low1) {
         low = low1;
     }
     printf("%d\n", large);
     printf("%d\n", low);
     return 0;
}

#include <stdio.h>

int max_of_four(int a,int b, int c, int d)
{
    int max=0;
    max = (a > b ? (a > c ? (a > d ?   a :  d) :  (c > d ? c : d) ) : (b > c ? (b > d ? b : d>c? d : c ) : c>d? c : d)) ;

   return max;
}

int main() {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int ans = max_of_four(a, b, c, d);
    printf("%d", ans);

    return 0;
}

下面是使用单个if语句的替代解决方案：

#include <stdio.h>

int main() {
    int a, b, c, d;
  
    printf("enter the numbers: ");
    if (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) {
        int lo1 = a * (a <= b) + (a > b) * b;
        int hi1 = a * (a >= b) + (a < b) * b;
        int lo2 = c * (c <= d) + (c > d) * d;
        int hi2 = c * (c >= d) + (c < d) * d;
        int lo = lo1 * (lo1 <= lo2) + (lo1 > lo2) * lo2;
        int hi = hi1 * (hi1 >= hi2) + (hi1 < hi2) * hi2;
        printf("%d %d\n", hi, lo);
        return 0;
    } else {
        return 1;
    }
}

这是一个很有创造性的问题。我只用了两个if语句就解决了你的问题。

int a = 22, b = 25, c = 100, d = 4;
        int max = 0, min = 0;
        if ((a > b ? (a > c ? (a > d ? max = a : max = d) : max = c) : (b > c ? (b > d ? max = b : max = d) : max = c)) == 1)
        {
              //Control never goes inside this loop
        }
        else if ((a < b ? (a < c ? (a < d ? min= a : min= d) : min= c) : (b < c ? (b < d ? min = b : min = d) : min = c)) == 1)
        {
             //Control never goes inside this loop
        }
        printf("Maximum Value is:%d and Minimum Value is: %d", max, min);//at this statement max and min will have appropriate values