如何使用C语言中的for循环将数组中的偶数和奇数分开？

数组是一组以单一名称存储的相关数据项。

例如 int Student[30]; //student是一个数组名，包含单个变量名的30个数据项集合

数组的操作

• 搜索 - 用于查找特定元素是否存在

• 排序 - 它有助于排列数组中的元素按升序或降序排列。

• 遍历 - 它按顺序处理数组中的每个元素。

• 插入 - 它有助于在数组中插入元素。

• 删除 - 它有助于删除数组中的元素。数组中的元素。

在数组中查找偶数的逻辑如下 -

for(i = 0; i < size; i ++){
   if(a[i] % 2 == 0){
      even[Ecount] = a[i];
      Ecount++;
   }
}

在数组中查找奇数的逻辑如下 -

for(i = 0; i < size; i ++){
   if(a[i] % 2 != 0){
      odd[Ocount] = a[i];
      Ocount++;
   }
}

要显示偶数，请调用下面提到的显示函数 -

printf("no: of elements comes under even are = %d 
", Ecount);
printf("The elements that are present in an even array is: ");
void display(int a[], int size){
   int i;
   for(i = 0; i < size; i++){
      printf("%d \t ", a[i]);
   }
   printf("
");
}

要显示奇数，请按照以下方式调用显示函数 −

printf("no: of elements comes under odd are = %d 
", Ocount);
printf("The elements that are present in an odd array is : ");
void display(int a[], int size){
   int i;
   for(i = 0; i < size; i++){
      printf("%d \t ", a[i]);
   }
   printf("
");
}

程序

以下是使用 for 循环分隔数组中偶数和奇数的 C 程序 -

现场演示

#include<stdio.h>
void display(int a[], int size);
int main(){
   int size, i, a[10], even[20], odd[20];
   int Ecount = 0, Ocount = 0;
   printf("enter size of array :
");
   scanf("%d", &size);
   printf("enter array elements:
");
   for(i = 0; i < size; i++){
      scanf("%d", &a[i]);
   }
   for(i = 0; i < size; i ++){
      if(a[i] % 2 == 0){
         even[Ecount] = a[i];
         Ecount++;
      }
      else{
         odd[Ocount] = a[i];
         Ocount++;
      }
   }
   printf("no: of elements comes under even are = %d 
", Ecount);
   printf("The elements that are present in an even array is: ");
   display(even, Ecount);
   printf("no: of elements comes under odd are = %d 
", Ocount);
   printf("The elements that are present in an odd array is : ");
   display(odd, Ocount);
   return 0;
}
void display(int a[], int size){
   int i;
   for(i = 0; i < size; i++){
      printf("%d \t ", a[i]);
   }
   printf("
");
}

输出

当执行上述程序时，会产生以下结果 -

enter size of array:
5
enter array elements:
23
45
67
12
34
no: of elements comes under even are = 2
The elements that are present in an even array is: 12 34
no: of elements comes under odd are = 3
The elements that are present in an odd array is : 23 45 67

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