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Week 17 free intrusion pointer exercise - output maximum

2022-07-26 17:46:00 【51CTO】

Problems and code ：

Problem A: Pointer practice -- Output Max

       
       Time Limit: 1 Sec  
       
 Memory Limit:  128 MB
       
Submit: 1912  
       
 Solved: 1212
       
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Description

notes ： This question only needs to submit the code of the part to be filled in

Using pointer method , Output 10 The maximum and minimum values of integer numbers

       
       /*C*/
       
#include<stdio.h>
       
 int main()
       
 {
       
     int a[10],i,max,min,*p;
       
     p=a;
       
     for(i=0; i<10; i++)
       
         scanf("%d",p++);
       
     max=min=a[0];
       
     for(p=a; p<a+10; p++)
       
       /*******************************
       
       Please supplement the missing code in this section 
       
      ********************************/     
       
     printf("max=%d\n",max);
       
     printf("min=%d\n",min);
       
     return 0;
       
 }
      
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Input

10 It's an integer

Output

Maximum and minimum

Sample Input

       
       2 6 3 8 1 5 7 0 4 9
      
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Sample Output

       
       max=9
       
min=0
      
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       /* School of computer science, Yantai University 
       
        author ： Jingyile 
       
        Completion time ：2016 year 12 month 22 Japan 
       
       */
       
       #include<stdio.h>
       
       int 
       
       main()
       
{
       
       int 
       
       a[
       
       10],
       
       i,
       
       max,
       
       min,
       
       *
       
       p;
       
       p
       
       =
       
       a;
       
       for(
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       10; 
       
       i
       
       ++)
       
       scanf(
       
       "%d",
       
       p
       
       ++);
       
       max
       
       =
       
       min
       
       =
       
       a[
       
       0];
       
       for(
       
       p
       
       =
       
       a; 
       
       p
       
       <
       
       a
       
       +
       
       10; 
       
       p
       
       ++)
       
    {
       
       if(
       
       *(
       
       p)
       
       >
       
       max)
       
       max
       
       =*
       
       p;
       
       if(
       
       *(
       
       p)
       
       <
       
       min)
       
       min
       
       =*
       
       p;
       
    }
       
       printf(
       
       "max=%d\n",
       
       max);
       
       printf(
       
       "min=%d\n",
       
       min);
       
       return 
       
       0;
       
}
      
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