奇数和偶数计数器

Question

我想更改此代码以从输入中获得20个数字并计算奇数的数量和偶数的数量？请任何人都可以帮忙？

#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

这里是一个解决您的问题的函数：

提示：您需要一个循环来把你输入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。现在去了解一个函数是什么。

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); 
}