Question

分别求出数组中所有奇数之和以及所有偶数之和。

1. 下列程序的功能是分别求出数组中所有奇数之和以及所有偶数之和。形参n给了数组中数据的个数，利用指针odd返回奇数之和，利用指针even返回偶数之和。请在下面空白处将实现这一功能的函数完善，并且调试运行出结果。#include <stdio.h>#define N 10void fun(int *a,int n,int *odd,int *even){ }void main(){ int a[N]={1,10,2,3,19,6},i,n=6,odd,even; printf("The original data is:\n"); for(i=0;i<n;i++) printf("%5d",*(a+i)); printf("\n\n"); fun(a,n,&odd,&even); printf("The sum of odd numbers:%d\n",odd); printf("The sum of even number:%d\n",even);

Answer 1

代码： #include <stdio.h>#define N 10void fun(int *a,int n,int *odd,int *even)
{
int i;
*even=0;
*odd=0;
for(i=0;i<n;i++)
{
if(a[i]%2==0)
*even=*even+a[i];
else
*odd=*odd+a[i];
}
}void main(){ int a[N]={1,10,2,3,19,6},i,n=6,odd,even; printf("The original data is:\n"); for(i=0;i<n;i++) printf("%5d",*(a+i)); printf("\n\n"); fun(a,n,&odd,&even); printf("The sum of odd numbers:%d\n",odd); printf("The sum of even number:%d\n",even);
} 图：