POTD 2026-07-15 solved

src/3600-3699/3658.cpp

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+/*
+Tags
+level-easy
+math, number-theory
+
+Problem Description
+3658. GCD of Odd and Even Sums
+
+You are given an integer n. Your task is to compute the GCD (greatest common divisor) of two values:
+sumOdd: the sum of the smallest n positive odd numbers.
+sumEven: the sum of the smallest n positive even numbers.
+Return the GCD of sumOdd and sumEven.
+
+Example 1:
+Input: n = 4
+Output: 4
+Explanation:
+Sum of the first 4 odd numbers sumOdd = 1 + 3 + 5 + 7 = 16
+Sum of the first 4 even numbers sumEven = 2 + 4 + 6 + 8 = 20
+Hence, GCD(sumOdd, sumEven) = GCD(16, 20) = 4.
+
+Example 2:
+Input: n = 5
+Output: 5
+Explanation:
+Sum of the first 5 odd numbers sumOdd = 1 + 3 + 5 + 7 + 9 = 25
+Sum of the first 5 even numbers sumEven = 2 + 4 + 6 + 8 + 10 = 30
+Hence, GCD(sumOdd, sumEven) = GCD(25, 30) = 5.
+
+Constraints:
+1 <= n <= 10​​​​​​​00
+*/
+
+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+#include <numeric>
+#include <cmath>
+#include <queue>
+#include <stack>
+#include <set>
+#include <map>
+#include <unordered_map>
+#include <unordered_set>
+#include <sstream>
+#include <limits>
+#include <iomanip>
+#include <functional>
+#include <cstring>
+#include <climits>
+#include "../core/core.h"
+using namespace std;
+
+class Solution
+{
+public:
+	int gcdOfOddEvenSums(int n)
+	{
+		int oddSum = (n * (2 * 1 + ((n - 1) * 2))) / 2;
+		int evenSum = (n * (2 * 2 + ((n - 1) * 2))) / 2;
+
+		return gcd(oddSum, evenSum);
+	}
+};
+
+int main()
+{
+	int n;
+	cin >> n;
+	Solution sol;
+	int result = sol.gcdOfOddEvenSums(n);
+	cout << result << "\n";
+
+	return 0;
+}

tests/3600-3699/3658.txt

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+Test Case 1:
+Input:
+4
+Output:
+4
+
+Test Case 2:
+Input:
+5
+Output:
+5