Project Euler #1: Multiples of 3 and 5 Discussions |

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2 weeks ago+ 0 comments

or anyone who is using Python3. If you've already converted from the iterive to the O(1) solution but you're still getting the "wrong answer" with Test Cases 2 and 3 here is why: When you divide integers in python 3, the answer is converted automatically to a float. Becuase Test Cases 2 and 3 involve very large numbers, when you convert back to integer, there will be a huge rounding error and your answers will be wrong. so rather than dividing with 2 or multiplying by 0.5 just do right shift by 1.

3 weeks ago+ 0 comments

I am getting Runtime Error for Test Case 2 & 3. The logic seems fine, can someone please help?

#!/bin/python3

import sys

t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    p = 0
    if 1 <= t <= 10 ** 5 and 1 <=n <= 10 ** 9:
        p = [x for x in range(1,n) if x % 3 == 0 or x % 5 == 0]
        print(sum(p))

4 weeks ago+ 0 comments

tried so hard to get this right :')

import sys
t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    a = ((n-1)//3)
    b = ((n-1)//5)
    c = ((n-1)//15)
    answer = (3*a*(a+1)//2) + (5*b*(b+1)//2) - (15*c*(c+1)//2)
    print (int(answer))

1 month ago+ 0 comments

If you approached the question using AP and have removed the repeating elements (i.e., sum of multiples of 15) from the: sum of AP of 3 + sum of AP of 5, then the only thing you need to make sure is that integer overflow does not occur. This is the reason why your testcases 2 & 3 might not pass. One trivial way to ensure that integer overflow doesn't occur is to use long long int for every variable. Another way is to type cast befor overflow occurs for storing the intermediate values. Here's my solution for your reference.

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        long long int n;
        cin >> n;
        long long int a = (n - 1) / 3;
        long long int b = (n - 1) / 5;
        long long int sum = a * (3 + 3 * a) >> 1;
        sum += b * (5 + 5 * b) >> 1;
        long long int c = (n - 1) / 15;
        sum -= c * (15 + 15 * c) >> 1;
        cout << sum << "\n";
    }
    return 0;	
		
}

1 month ago+ 0 comments

import sys
t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    three=(n-1)//3
    five=(n-1)//5
    fifteen=(n-1)//15
    print((3*(three*(three+1)//2)+5*(five*(five+1)//2)-15*(fifteen*(fifteen+1)//2)))

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