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
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 </p><p>", 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("</p><p>"); }
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 </p><p>", 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("</p><p>"); }
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 :</p><p>"); scanf("%d", &size); printf("enter array elements:</p><p>"); 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 </p><p>", Ecount); printf("The elements that are present in an even array is: "); display(even, Ecount); printf("no: of elements comes under odd are = %d </p><p>", 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("</p><p>"); }
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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