Question

我自学C和我试图制作2个锻炼计划：

第一个拿一个数字并检查它是偶数还是奇数; 这是我为第一个提出的：

#include <stdio.h>
int main(){
      int n;
      printf("Enter a number that you want to check: ");
      scanf("%d",&n);
      if((n%2)==0)      
           printf("%d is even.",n);
      else
           printf("%d is odd.",n);
      return 0;
}

第二个应该将 n 数字作为输入，并计算输入的数字中的偶数，奇数和零的数量。输出应该是偶数，奇数和零的数量。

我想问一下如何在这种情况下实现循环：如果每个整数都可以接受，我怎么能设置一个EOF值（所以我不能，比方说，把0结束）？你能告诉我如何有效地构建这个短代码吗？

Answer 1

#include <stdio.h>
int main(void) {
    int n, nEven=0, nOdd=0, nZero=0;

    for (;;) {
        printf("\nEnter a number that you want to check: ");
        //Pressing any non-numeric character will break;
        if (scanf("%d", &n) != 1) break;

        if (n == 0) {
            nZero++;
        }
        else {
            if (n % 2) {
                nEven++;
            }
            else {
                nOdd++;
            }
        }
    }

    printf("There were %d even, %d odd, and %d zero values.", nEven, nOdd, nZero);
    return 0;
}

enter image description here

Answer 2

检查scanf()

的返回值

1,1填充字段（n）填写0,0字段，可能为数字输入了类似“abc”的字符。
EOF，遇到文件结尾（或很少出现IO错误）。

#include <stdio.h>
int main(void) {
  int n;

  for (;;) {  
    printf("Enter a number that you want to check: ");
    if (scanf("%d",&n) != 1) break;
    if((n%2)==0)      
      printf("%d is even.",n);
    else
      printf("%d is odd.",n);
  }
  return 0;
}

或读取随后阅读的数字计数：

int main(void) {
  int n;

  printf("Enter the count of numbers that you want to check: ");
  if (scanf("%d",&n) != 1) Handle_Error();

  while (n > 0) {  
    n--;
    printf("Enter a number that you want to check: ");
    int i;
    if (scanf("%d",&i) != 1) break;
    if((i%2)==0) {
      if (i == 0) printf("%d is zero.\n",i);
      else printf("%d is even and not 0.\n",i);
    }
    else
      printf("%d is odd.\n",i);
  }
  return 0;
}

Answer 3

嘿，看看这个

#include<stdio.h>
#include<conio.h>

void main()

{    

       int nodd,neven,num,digit ; 
       clrscr();



       printf("Count number of odd and even digits in a given integer number ");

       scanf("%d",&num);

                nodd = neven =0; /* count of odd and even digits */

                while (num> 0)

                    {

                        digit = num % 10; /* separate LS digit from number */

                        if (digit % 2 == 1)

                           nodd++;

                        else neven++;

                               num /= 10; /* remove LS digit from num */

                    }

                        printf("Odd digits : %d Even digits: %d\n", nodd, neven);

                        getch();

}

Answer 4

您可以这样做：

#include <stdio.h>
int main(){
    int n,evenN=0,oddN=0,zeros=0;
    char key;
    do{
        clrscr();
        printf("Enter a number that you want to check: ");
        scanf("%d",&n);
        if(n==0){
            printf("%d is zero.",n);
            zeros++;
        }
        else if((n%2)==0){      
            printf("%d is even.",n);
            evenN++; 
        }
        else{
            printf("%d is odd.",n);
            oddN++;
        }
        puts("Press ENTER to enter another number. ESC to exit");
        do{
        key = getch();
        }while(key!=13 || key!=27) //13 is the ascii code fore enter key, and 27 is for escape key

    }while(key!=27) 
    clrscr();
    printf("Total even numbers: %d",evenN);
    printf("Total odd numbers: %d",oddN);
    printf("Total odd numbers: %d",zeros);
    return 0;
}

此程序要求输入一个号码，评估号码，然后要求继续拨打另一个号码或退出。

计算偶数和奇数的程序

4 个答案: