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  1. #include<stdio.h>
  2. #define SIZE 5
  3.  int Getmax(int *ac,int n)
  4. {
  5. 	int i,max;
  6. 	for(i=0;i<=2;i++)
  7. 	{
  8. 		max = (ac[i]>ac[i+1]?ac[i]:ac[i+1]);
  9. 		max =(max>ac[i+2]?max:ac[i+2]);
  10.  
  11. 	}
  12. 	return max;
  13.  
  14. }
  15.  
  16.  int Getmin(int *ar,int m)
  17. {
  18. 	int i,min;
  19. 	for(i=0;i<=2;i++)
  20. 	{
  21. 		min = (ar[i]>ar[i+1]?ar[i+1]:ar[i]);
  22. 		min =(min>ar[i+2]?ar[i+2]:min);
  23. 	}
  24. 	return min;
  25. }
  26.  
  27.  
  28.  
  29. int main(void)
  30. {
  31. 	 int a[SIZE],temp1,temp2,i;
  32.  
  33. 	printf("请输入5个范围在10000以内的整数：");
  34. 	for(i=0;i<=4;i++)
  35. 	{
  36. 		scanf("%d",&a[i]);
  37. 	}
  38. 	temp1=Getmin(a,SIZE);
  39. 	temp2=Getmax(a,SIZE);
  40. 	printf("max=%d,min=%d",temp2,temp1);
  41.  
  42. }
Success #stdin #stdout 0s 3100KB
comments (?)
stdin 
Standard input is empty
stdout 
请输入5个范围在10000以内的整数：max=134518936,min=-1217105464
https://ideone.com/gm4kqE
language:
C++ (gcc 8.3)
created:
10 years ago
visibility:
public

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