C中偶数位和奇数位的数字总和

Question

我需要在整数的偶数和奇数位置添加数字。 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

这是您的问题的解决方案：

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);
}

Answer 2

好的，让我们首先假设该数字有偶数位数，因此倒数第二和最后一个分别位于奇数和偶数位置。

然后，可以使用number % 10检索最后一位数字，并使用(number / 10) % 10检索倒数第二个(number / 10) % 10 。

因此，知道这一点后，您可以简单地遍历该数字，将这些值相加并除以 100，直到得到一个小于 10 的数字。

如果该数字为零，那么您对原始数字为偶数的假设是正确的，您可以退出。

如果它是非零（ 1..9 ），那么您的假设是错误的，您应该交换偶数和奇数和，然后将最后一位数字添加到奇数和。

伪代码是这样的：

def getEvenOdd (num):
    even = 0
    odd = 0

    while num > 9:
        even = even + (num % 10)
        odd = odd + ((num / 10) % 10)
        num = num / 100

    if num > 0:
        temp = even
        even = odd
        odd = temp
        odd = odd +  num

    return (even,odd)

print getEvenOdd(1234567)

其输出将类似于(12,16) 。

---

伪代码看起来像初学者的 Python 绝非偶然，因为该语言是完美的伪代码语言，前提是您不要弄乱它的更复杂的角落。

Answer 3

检查以下代码：

#include <stdio.h>

int main(void){
char p[20];
int i=0,even=0,odd=0,n;
scanf("%d",&n);
sprintf(p,"%d",n);
while(p[i] != '\0')
{
    if(i%2 == 0)
    odd+= (p[i] - '0');
    else
    even+= (p[i] - '0');

    i++;
}
printf("%d\n",even);
printf("%d\n",odd);
return 0;
}

Answer 4

我的五美分。:)

#include <stdio.h>

typedef struct pair
{
    unsigned int odd;
    unsigned int even;
} pair_t;   

pair_t GetSums( unsigned int x )
{
    const unsigned int Base = 10;
    pair_t sums = { 0u, 0u };
    size_t n = 0;

    do
    {
        if ( ++n & 1 ) sums.odd += x % Base;
        else sums.even += x % Base;
    } while ( x /= Base );

    if ( ( n & 1 )  == 0 )
    {
        unsigned int tmp = sums.odd;
        sums.odd = sums.even;
        sums.even = tmp; 
    }

    return sums;
}

int main(void) 
{
    while ( 1 )
    {
        unsigned int x;
        pair_t sums;

        printf( "\nEnter a non-negative number (0 - exit): " );
        if ( scanf( "%u", &x ) != 1 || x == 0 ) break;

        sums = GetSums( x );

        printf( "\nSum of digits in odd  positions is %u\n", sums.odd );            
        printf( "Sum of digits in even positions is %u\n", sums.even );         

    }

    return 0;
}

输出可能看起来像

Enter a non-negative number (0 - exit): 1234567

Sum of digits in odd  positions is 16
Sum of digits in even positions is 12

Enter a non-negative number (0 - exit): 123456

Sum of digits in odd  positions is 9
Sum of digits in even positions is 12

Enter a non-negative number (0 - exit): 0

Answer 5

要对所有其他数字求和，只需在每次迭代中除以 100：

int sumodd(int num) {
    int sum = 0, rem;
    while(num) {
        rem=num%10;
        sum=sum+rem;
        num=num/100);
    }
    return sum;
}

由于我们已经有了一个可以对所有其他数字求和的函数，我们可以重用它：

int sumeven(int num) {
    return sumodd(num/10);
}