Question

我想更改此代码以从输入中获取20个数字，并计算有多少是奇数，有多少是偶数？请有人帮忙吗？！

#include <conio.h>
#include <stdio.h>
int main()
{
 int n;
 int odd=0;
 int even=0;
  printf("\nEnter any number \n");
  scanf("%d",&n);    
 if(n%2!=0)
 {
     printf("%d is an odd number",n);
     odd++;
 }
 else
 {
     printf("%d is an even number",n);
     even++;
 }
printf("\n odd%d / even%d",odd,even);
}

Answer 1

这是一个解决问题的函数：

提示：您需要一个循环来输入您的输入20次。

void countForJHikaam(){
    int n,i;
    int odd=0;
    int even=0;

    for(i=0;i<20;i++){

        scanf("%d\n",&n);
        if(n%2==0){
            even++;
        }else{odd++;}

    }
    printf("Odds: %d, Evens: %d",odd,even);
}

它不会真正帮助你学习。现在去了解功能是什么。

Answer 2

#include <conio.h>
#include <stdio.h>
int main()
{
   int n;
   int odd=0;
   int even=0;
   printf("\nEnter any number \n");

   while(scanf("%d",&n))    
     (n%2) ? (++odd) : (++even);

   printf("\n odd%d / even%d",odd,even);
}

Answer 3

#include<stdio.h>
main()
{
     int odd=0,even=0,no,count=20;
     printf("Enter the 20 numbers...\n");
here:
     scanf("%d",&no);
     (no%2==0)? odd++ : even++ ;`
     count--;
     if(count>0)
     goto here;

     printf("No of odd numbers... :%d\n",odd);
     printf("No of even numbers... :%d\n",even);
}

奇怪甚至反击

3 个答案: