题解 | 3和5的倍数

#include <iostream>
using namespace std;

class Solution {
public:
    int sumOfMultiples(int n) {
        // 将n减1，因为我们要求小于n的数
        n--;
        
        // 计算3的倍数之和
        int count3 = n / 3;
        int sum3 = 3 * (count3 * (count3 + 1)) / 2;
        
        // 计算5的倍数之和
        int count5 = n / 5;
        int sum5 = 5 * (count5 * (count5 + 1)) / 2;
        
        // 计算15的倍数之和（被重复计算的部分）
        int count15 = n / 15;
        int sum15 = 15 * (count15 * (count15 + 1)) / 2;
        
        // 返回总和
        return sum3 + sum5 - sum15;
    }
};

int main() {
    int n;
    cin >> n;
    
    Solution solution;
    cout << solution.sumOfMultiples(n) << endl;
    
    return 0;
}

import java.util.Scanner;

public class Main {
    static class Solution {
        public int sumOfMultiples(int n) {
            // 将n减1，因为我们要求小于n的数
            n--;
            
            // 计算3的倍数之和
            int count3 = n / 3;
            int sum3 = 3 * (count3 * (count3 + 1)) / 2;
            
            // 计算5的倍数之和
            int count5 = n / 5;
            int sum5 = 5 * (count5 * (count5 + 1)) / 2;
            
            // 计算15的倍数之和（被重复计算的部分）
            int count15 = n / 15;
            int sum15 = 15 * (count15 * (count15 + 1)) / 2;
            
            // 返回总和
            return sum3 + sum5 - sum15;
        }
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        Solution solution = new Solution();
        System.out.println(solution.sumOfMultiples(n));
        
        sc.close();
    }
}

class Solution:
    def sum_of_multiples(self, n: int) -> int:
        # 将n减1，因为我们要求小于n的数
        n -= 1
        
        # 计算3的倍数之和
        count3 = n // 3
        sum3 = 3 * (count3 * (count3 + 1)) // 2
        
        # 计算5的倍数之和
        count5 = n // 5
        sum5 = 5 * (count5 * (count5 + 1)) // 2
        
        # 计算15的倍数之和（被重复计算的部分）
        count15 = n // 15
        sum15 = 15 * (count15 * (count15 + 1)) // 2
        
        # 返回总和
        return sum3 + sum5 - sum15

def main():
    n = int(input())
    solution = Solution()
    print(solution.sum_of_multiples(n))

if __name__ == "__main__":
    main()