10,000以下的数字总和，是Java中3,5或7的倍数

Question

I know how to get the program to add up the sums of the multiple for each of 3, 5 and 7, but I'm not sure how I'd get the program to only use each number once. For example, I can get the program to find out all of the numbers and add them up for 3 and then do the same for 5, but then the number 15 would be in the final number twice. I'm not sure exactly how I'd get it to only take the number once. Thanks for any help.

Answer 1

While the generate-and-test approach is simple to understand, it is also not very efficient if you want to run this on larger numbers. Instead, we can use the inclusion-exclusion principle .

The idea is to first sum up too many numbers by looking at the multiples of 3, 5 and 7 separately. Then we subtract the ones we counted twice, ie multiples of 3*5, 3*7 and 5*7. But now we subtracted too much and need to add back the multiples of 3*5*7 again.

We start by finding the sum of all integers 1..n which are multiples of k . First, we find out how many there are, m = n / k , rounded down thanks to integer division. Now we just need to sum up the sequence k + 2*k + 3*k + ... + m*k . We factor out the k and get k * (1 + 2 + ... + m) .

This is a well-known arithmetic series, which we know sums to k * m * (m + 1)/2 (See triangle number ).

private long n = 9999;
private long multiples(long k) {
    long m = n / k;
    return k * m * (m + 1) / 2:
}

Now we just use inclusion-exclusion to get the final sum:

long sum = multiples(3) + multiples(5) + multiples(7)
         - multiples(3*5) - multiples(3*7) - multiples(5*7)
         + multiples(3*5*7);

This will scale much better to larger n than just looping over all the values, but beware of overflow and change to BigInteger s if necessary.

Answer 2

The easiest approach would be to use a for loop thus:

int sum = 0;
for(int i=1; i<10000; i++)
{
    if (i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
        sum += i;
}

Answer 3

使用Set存储唯一的倍数，然后对Set的值求和。

Answer 4

I would use a Set . This way you are guaranteed that you won't get any duplicates if they are your main problem.

Answer 5

One simple solution would be to add each number thats a multiple of 3,5, or 7 to an Answer list. And then as you work thru each number, make sure that its not already in the answer list.

(pseudo-code)

List<int> AnswerList;
List<int> MultiplesOfFive;
List<int> MultiplesOfSeven;
List<int> MultiplesOfThree;

for (int i = 0 ; i < 10000; i++)
{
  if ( i % 3 == 0 && AnswserList.Contains(i) == false)
  {
    MultiplesOfThree.Add(i); 
    AnswerList.Add(i);
  }
  if ( i % 5 == 0 && AnswserList.Contains(i) == false)
  {
    MultiplesOfFive.Add(i); 
    AnswerList.Add(i);
  }
  if ( i % 7 == 0 && AnswserList.Contains(i) == false)
  {
    MultiplesOfSeven.Add(i); 
    AnswerList.Add(i);
  }
}

Answer 6

对于循环1到1000使用i <= 10000的解决方案，否则它将跳过10000本身，这是5的倍数。道歉，由于某种原因我不能发布这个作为评论