c - C中偶数位和奇数位的数字之和

Question

我需要将整数中偶数和奇数位置的数字相加。假设数字 = 1234567。偶数位数字之和 = 2+4+6 = 12 奇数位数字之和 = 1+3+5+7 = 16

我目前拥有的代码是:

int returnsum(int num) {
  while (num) {
    rem = num % 10;
    sum = sum + rem;
    num = num / 10;
  }
  while (sum) {
    a = sum % 10;
    arr[i] = a:
      sum = sum / 10;
    i++;
  }
  for (i = 0; a[i]; i += 2) {
    summ = summ + a[i];
  }
  return summ;
}

void main() {
  int n, m, oddSum = 0, evenSum = 0;
  printf("Please insert the number for the program:");
  scanf("%d", & n);
  while (n != 0) {
    oddSum += n % 10;
    n /= 10;
    evenSum += n % 10;
    n /= 10;
  }
  printf("Sum of digits in even  places:%d\n", evenSum);
  printf("Sum of digits in odd  places:%d\n", oddSum);
}

Answer 1

这是您问题的解决方案:

void main()
{
 int n,m,oddSum=0,evenSum=0;
 printf("Please insert the number for the program:");
 scanf("%d",&n);
 int flag=0;
 int counter=1;
 while (n!=0) {
   if(counter%2==0)
   {
     evenSum += n % 10;
     n /= 10;
   }
   else
   {
     oddSum += n % 10;
     n /= 10;
   }
   counter++;
 }

 if(counter%2==0)
 {
   int temp=oddSum;
   oddSum=evenSum;
   evenSum=temp;
 }
 printf("Sum of digits in even  places:%d\n",evenSum);
 printf("Sum of digits in odd  places:%d\n",oddSum);
}