N以下3或5的所有倍数之和

Question

I have to find the sum of all the multiples of 3 or 5 below N. example if we have to list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9,the sum of these multiples is 23.

Now the only problem left is that I want it to be able to read ALL the numbers THEN display the sum, for now it reads one number and display the sum right after it, any ideas?

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Solution{
public static void main(String[] args) throws IOException {


    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String line = br.readLine();
    int Nbr = Integer.parseInt(line);


        for(int j=0; j<Nbr;j++)
        {
            BufferedReader br2 = new BufferedReader(new   InputStreamReader(System.in));
            String line2 = br2.readLine();
            String[] numbers = new String[Nbr];
            numbers[j]= line2;
            System.out.println(Somme(Long.parseLong(numbers[j])));
        }

        }


public static long Somme(long Nn) {
    long s = 0;
    for (int i = 0; i < Nn; i++) {
        if (((i % 3) == 0) || ((i % 5) == 0)) {
            s = s + i;
        }
    }
    return (s);
}
}

Answer 1

I suggest you use a Scanner . I would also suggest you add to a sum by iterating from 3 to n in increments of 3 . Then from 5 to n in increments of 5 (then exclude the multiples of 3 , because they've already been added). Something like

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    while (scanner.hasNextInt()) {
        int n = scanner.nextInt();
        System.out.println(getSum(n));
    }
}

public static long getSum(int n) {
    long sum = 0;
    for (int i = 3; i < n; i += 3) {
        sum += i;
    }
    for (int i = 5; i < n; i += 5) {
        if (i % 3 != 0) { // <-- already added if i is divisible by 3
            sum += i;
        }
    }
    return sum;
}

Based on your comment below, change main to first read the int of counts, then store them in an array. Something like

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    while (scanner.hasNextInt()) {
        int toRead = scanner.nextInt();
        int[] vals = new int[toRead];
        for (int t = 0; t < toRead; t++) {
            if (scanner.hasNextInt()) {
                vals[t] = scanner.nextInt();
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int n : vals) {
            sb.append(getSum(n)).append(" ");
        }
        System.out.println(sb);
    }
}

Answer 2

Move your print statement outside the loop.

Before loop,

long[] sumArr = new long[Nbr];

inside loop,

sumArr[j] = Somme(Long.parseLong(numbers[j]));

after loop,

for (long sum : sumArr) {
    System.out.println(sum);
}

Answer 3

如果您想先读取所有N，请先读取它们并将它们放入数组中，然后尝试求和

Answer 4

您可以制作ArrayList并将其总和添加到其中，然后在循环后将其打印出来

Answer 5

As we know that we need to sum of those numbers which are divisible by 3 or 5 below n where n=10 or 100 or more (it is limit). But if we have n=20 so 15(15%5==0 && 15%3==0) will come twice so that we need to add only once so that we need to check a number is divisible by 15(since 5*3=15).

   long n1=0,n2=0;
    int i=1,j=1;
    while(3*i<n)
        {
        n1 +=3*i;
        i++;
        if(5*j<n)
            {
            n2+=5*j;
            if(5*j%15==0)
                n2 -=5*j;
            j++;
        }
      }

    System.out.println(n1+n2);
    }