C 语言查找数组中奇偶数之和的程序

如何使用 For 循环、While 循环、函数编写 C 语言程序来查找数组中奇偶数的和，并附带示例。

C 语言查找数组中奇偶数之和的程序

此程序允许用户输入一维数组的大小和行元素。

/* C Program to find Sum of Even and Odd Numbers in an Array */
#include<stdio.h>
 
int main()
{
 int Size, i, a[10];
 int Even_Sum = 0, Odd_Sum = 0;
 
 printf("\n Please Enter the Size of an Array : ");
 scanf("%d", &Size);
 
 printf("\nPlease Enter the 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_Sum = Even_Sum + a[i];
   }
   else
   {
 	Odd_Sum = Odd_Sum + a[i];
   }
 }
  
 printf("\n The Sum of Even Numbers in this Array = %d ", Even_Sum);
 printf("\n The Sum of Odd Numbers in this Array = %d ", Odd_Sum);
 return 0;
}

在这里，For 循环将确保数字在 0 和最大值之间。在此示例中，它将从 0 到 4。

for(i = 0; i < Size; i ++)

在下一行，我们声明了If语句

if(a[i] % 2 == 0)

任何能被 2 整除的数都是偶数。If 语句将检查当前 一维数组元素除以 2 的余数是否恰好等于 0。

• 如果条件为真，则为偶数，编译器将把数组元素添加到 Even_Sum。
• 如果条件为假，则为奇数。然后编译器将把数组元素添加到 Odd_Sum。

使用 While 循环查找数组中奇偶数之和的程序

这个程序与上面的相同，但这次我们使用了 While 循环来对数组中的奇偶数进行求和。

/* C Program to find Sum of Even and Odd Numbers in an Array */
#include<stdio.h>
int main()
{
 int Size, i, j = 0, a[10];
 int Even_Sum = 0, Odd_Sum = 0;
 
 printf("\n Please Enter the Size of an Array : ");
 scanf("%d", &Size);
 
 printf("\nPlease Enter the Array Elements\n");
 for(i = 0; i < Size; i++)
 {
      scanf("%d", &a[i]);
 } 
 while(j < Size)
 {
   if(a[j] % 2 == 0)
   {
 	Even_Sum = Even_Sum + a[j];
   }
   else
   {
 	Odd_Sum = Odd_Sum + a[j];
   }
   j++;
 } 
 printf("\n The Sum of Even Numbers in this Array = %d ", Even_Sum);
 printf("\n The Sum of Odd Numbers in this Array = %d ", Odd_Sum);
 return 0;
}

Please Enter the Size of an Array : 6

Please Enter the Array Elements
14 26 53 19 89 156

 The Sum of Even Numbers in this Array = 196 
 The Sum of Odd Numbers in this Array = 161 

使用函数查找数组中奇偶数之和的程序

这个程序与第一个C 编程示例相同。但是，我们使用函数将查找偶数之和与查找奇数之和的逻辑分开了。

/* C Program to Count Even and Odd Numbers in an Array */
#include<stdio.h>

int CountEvenNumbers(int a[], int Size);
int CountOddNumbers(int a[], int Size);

int main()
{
 int Size, i, a[10];
 int Even_Count = 0, Odd_Count = 0;
 
 printf("\n Please Enter the Size of an Array\n");
 scanf("%d", &Size);
 
 printf("\nPlease Enter the Array Elements :  ");
 for(i = 0; i < Size; i++)
 {
      scanf("%d", &a[i]);
 }
 
 Even_Count = CountEvenNumbers(a, Size);
 Odd_Count = CountOddNumbers(a, Size);
  
 printf("\n Total Number of Even Numbers in this Array = %d ", Even_Count);
 printf("\n Total Number of Odd Numbers in this Array = %d ", Odd_Count);
 return 0;
}
int CountEvenNumbers(int a[], int Size)
{
  int i, Even_Count = 0;
  printf("\n List of Even Numbers in this Array: ");

  for(i = 0; i < Size; i ++)
  {
     if(a[i] % 2 == 0)
     {
 	printf("%d  ", a[i]);
 	Even_Count++;
     }
   }
   return Even_Count;
}
int CountOddNumbers(int a[], int Size)
{
  int i, Odd_Count = 0;
  printf("\n List of Odd Numbers in this Array: ");

  for(i = 0; i < Size; i ++)
  {
     if(a[i] % 2 != 0)
     {
 	printf("%d  ", a[i]);
 	Odd_Count++;
     }
   }
   return Odd_Count;
}

 Please Enter the Size of an Array
10

Please Enter the Array Elements :  10 15 25 69 78 96 47 57 94 109

 List of Even Numbers in this Array: 10  78  96  94  
 List of Odd Numbers in this Array: 15  25  69  47  57  109  
 Total Number of Even Numbers in this Array = 4 
 Total Number of Odd Numbers in this Array = 6