POTD 2026-07-16 solved

src/3800-3899/3867.cpp

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+/*
+Tags
+level-medium
+array, math, two-pointers, sorting, simulation, number-theory
+
+Problem Description
+3867. Sum of GCD of Formed Pairs
+
+You are given an integer array nums of length n.
+Construct an array prefixGcd where for each index i:
+Let mxi = max(nums[0], nums[1], ..., nums[i]).
+prefixGcd[i] = gcd(nums[i], mxi).
+After constructing prefixGcd:
+Sort prefixGcd in non-decreasing order.
+Form pairs by taking the smallest unpaired element and the largest unpaired element.
+Repeat this process until no more pairs can be formed.
+For each formed pair, compute the gcd of the two elements.
+If n is odd, the middle element in the prefixGcd array remains unpaired and should be ignored.
+Return an integer denoting the sum of the GCD values of all formed pairs.
+
+The term gcd(a, b) denotes the greatest common divisor of a and b.
+
+Example 1:
+Input: nums = [2,6,4]
+Output: 2
+Explanation:
+Construct prefixGcd:
+i	nums[i]	mxi	prefixGcd[i]
+0	2	2	2
+1	6	6	6
+2	4	6	2
+prefixGcd = [2, 6, 2]. After sorting, it forms [2, 2, 6].
+Pair the smallest and largest elements: gcd(2, 6) = 2. The remaining middle element 2 is ignored. Thus, the sum is 2.
+
+Example 2:
+Input: nums = [3,6,2,8]
+Output: 5
+Explanation:
+Construct prefixGcd:
+i	nums[i]	mxi	prefixGcd[i]
+0	3	3	3
+1	6	6	6
+2	2	6	2
+3	8	8	8
+prefixGcd = [3, 6, 2, 8]. After sorting, it forms [2, 3, 6, 8].
+Form pairs: gcd(2, 8) = 2 and gcd(3, 6) = 3. Thus, the sum is 2 + 3 = 5.
+
+Constraints:
+1 <= n == nums.length <= 10^5
+1 <= nums[i] <= 10​​​​​​^​9
+*/
+
+#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:
+	long long gcdSum(vector<int>& nums)
+	{
+		int n = nums.size();
+		vector<int> prefixGcd(n, 0);
+		int maxVal = nums[0];
+		for (int i = 0; i < n; i++)
+		{
+			maxVal = max(maxVal, nums[i]);
+			prefixGcd[i] = gcd(nums[i], maxVal);
+		}
+
+		sort(prefixGcd.begin(), prefixGcd.end());
+		int left = 0;
+		int right = n - 1;
+		long long result = 0;
+		while (left < right)
+		{
+			result += gcd(prefixGcd[left], prefixGcd[right]);
+			left++;
+			right--;
+		}
+
+		return result;
+	}
+};
+
+int main()
+{
+	int n;
+	cin >> n;
+	vector<int> nums(n, 0);
+	for (int i = 0; i < n; i++)
+	{
+		cin >> nums[i];
+	}
+	Solution sol;
+	long long result = sol.gcdSum(nums);
+	cout << result << "\n";
+	return 0;
+}

tests/3800-3899/3867.txt

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+Test Case 1:
+Input:
+3
+2 6 4
+Output:
+2
+
+Test Case 2:
+Input:
+4
+3 6 2 8
+Output:
+5