如何通过使用C语言中的for循环来分隔数组中的偶数和奇数?
数组是一组以单个名称存储的相关数据项。
例如,int学生[30];// student是一个数组名称,其中包含30个带有单个变量名称的数据项集合
数组的运算
搜索-用于查找是否存在特定元素
排序-有助于按升序或降序排列数组中的元素。
遍历-顺序处理数组中的每个元素。
插入-这有助于将元素插入数组中。
删除-它有助于删除数组中的元素。
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,请按如下所述调用显示功能-
printf("no: of elements comes under even are = %d \n", 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("\n");
}
要显示奇数,请调用显示功能,如下所示:
printf("no: of elements comes under odd are = %d \n", 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("\n");
}
程序
以下是C语言程序通过使用for循环来分隔数组中的偶数和奇数-
#include<stdio.h>输出结果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 :\n");
scanf("%d", &size);
printf("enter array elements:\n");
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 \n", Ecount);
printf("The elements that are present in an even array is: ");
display(even, Ecount);
printf("no: of elements comes under odd are = %d \n", 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("\n");
}
执行以上程序后,将产生以下结果-
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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