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  1. All Contests
  2. ProjectEuler+
  3. Project Euler #1: Multiples of 3 and 5
  4. Discussions

Project Euler #1: Multiples of 3 and 5

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1448 Discussions

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  • shyamganeshks
     + 0 comments
    long long sum_of_multiples(int x, int n) {
    int p = (n - 1) / x; // Largest multiple of x below n
    return (long long)x * p * (p + 1) / 2; // Sum of multiples formula
}
int main(){
    int t; 
    scanf("%d",&t);
    for(int a0 = 0; a0 < t; a0++){
        int n; 
        scanf("%d",&n);
        // Sum of multiples of 3 and 5, minus multiples of 15 (to avoid double counting)
        long long sum = sum_of_multiples(3, n) + sum_of_multiples(5, n) - sum_of_multiples(15, n);

        printf("%lld\n", sum);
    }
    return 0;
}

  • hmdmlkv
     + 0 comments

    I cannot pass 2 of the tests with this Kotlin code, anybody has any idea?

    fun main(args: Array<String>) {
    val t = readLine()!!.trim().toInt()

    for (tItr in 1..t) {
        val n = readLine()!!.trim().toInt()
        println("${sumn(n,3)+sumn(n,5)-sumn(n,15)}")
        
    }
    
}

fun sumn(n:Int,d:Int):Int{
    val l = (n-1)/d
    return d*l*(l+1)/2
}

  • satendrasharma12
     + 3 comments
    time limit problem in this program ,help to solve the problem

    int t; cin >> t; for(int a0 = 0; a0 < t; a0++){ int sum =0; int n; cin >> n; for(int i = 3; i < n; i++){ if(i%3==0 || i%5==0){ sum+=i; } }

        cout<<sum<<endl;
}

  • satendrasharma12
     + 0 comments

    int t; cin >> t; for(int a0 = 0; a0 < t; a0++){ int sum =0; int n; cin >> n; for(int i = 3; i < n; i++){ if(i%3==0 || i%5==0){ sum+=i; } }

        cout<<sum<<endl;
}

    time limit problem in this program ,help to solve the problem





hgdshfhs

  • mdhanifkhan116
     + 0 comments

    JAVA 8: Used sum of natural formula for calculation

    import java.util.Scanner;
public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        for(int i = 0; i < n; i++)
            { 
                long a = sc.nextLong();
                long x= (a-1)/3;
                long y= (a-1)/5;
                long z= (a-1)/15;
                
                long sum= (3*((x*(x+1))/2)) + (5*((y*(y+1))/2)) 
                        - (15*((z*(z+1))/2));
                
                System.out.println(sum);
            }
        sc.close();
        
     
}
}