feat: add solutions to lc problem: No.3312

solution/3300-3399/3312.Sorted GCD Pair Queries/README.md

@@ -84,32 +84,145 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3312.So
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：预处理 + 前缀和 + 二分查找
+
+我们可以预处理得到数组 $\textit{nums}$ 中的所有数对的最大公约数的出现次数，记录在数组 $\textit{cntG}$ 中。然后，我们计算数组 $\textit{cntG}$ 的前缀和。最后，对于每个查询，我们可以通过二分查找在数组 $\textit{cntG}$ 中找到第一个大于 $\textit{queries}[i]$ 的元素的下标，即为答案。
+
+我们用 $\textit{mx}$ 表示数组 $\textit{nums}$ 中的最大值，用 $\textit{cnt}$ 记录数组 $\textit{nums}$ 中每个数的出现次数。我们用 $\textit{cntG}[i]$ 表示数组 $\textit{nums}$ 中最大公约数等于 $i$ 的数对个数。为了计算 $\textit{cntG}[i]$，我们可以按照以下步骤进行：
+
+1. 计算数组 $\textit{nums}$ 中 $i$ 的倍数的出现次数 $v$，那么从这些元素中任选两个元素组成的数对一定满足最大公约数是 $i$ 的倍数，即 $\textit{cntG}[i]$ 需要增加 $v \times (v - 1) / 2$；
+1. 我们需要排除最大公约数是 $i$ 的倍数且大于 $i$ 的数对，因此，对于 $i$ 的倍数 $j$，我们需要减去 $\textit{cntG}[j]$。
+
+以上需要我们从大到小遍历 $i$，这样才能保证我们在计算 $\textit{cntG}[i]$ 时已经计算了所有的 $\textit{cntG}[j]$。
+
+最后，我们计算数组 $\textit{cntG}$ 的前缀和，然后对于每个查询，我们可以通过二分查找在数组 $\textit{cntG}$ 中找到第一个大于 $\textit{queries}[i]$ 的元素的下标，即为答案。
+
+时间复杂度 $O(n + (M + q) \times \log M)$，空间复杂度 $O(M)$。其中 $n$ 和 $M$ 分别是数组 $\textit{nums}$ 的长度和最大值，而 $q$ 是查询的数量。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:
+        mx = max(nums)
+        cnt = Counter(nums)
+        cnt_g = [0] * (mx + 1)
+        for i in range(mx, 0, -1):
+            v = 0
+            for j in range(i, mx + 1, i):
+                v += cnt[j]
+                cnt_g[i] -= cnt_g[j]
+            cnt_g[i] += v * (v - 1) // 2
+        s = list(accumulate(cnt_g))
+        return [bisect_right(s, q) for q in queries]
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int[] gcdValues(int[] nums, long[] queries) {
+        int mx = Arrays.stream(nums).max().getAsInt();
+        int[] cnt = new int[mx + 1];
+        long[] cntG = new long[mx + 1];
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i > 0; --i) {
+            int v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1L * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        int m = queries.length;
+        int[] ans = new int[m];
+        for (int i = 0; i < m; ++i) {
+            ans[i] = search(cntG, queries[i]);
+        }
+        return ans;
+    }
+
+    private int search(long[] nums, long x) {
+        int n = nums.length;
+        int l = 0, r = n;
+        while (l < r) {
+            int mid = l + r >> 1;
+            if (nums[mid] > x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
+        int mx = ranges::max(nums);
+        vector<int> cnt(mx + 1);
+        vector<long long> cntG(mx + 1);
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i; --i) {
+            long long v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1LL * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        vector<int> ans;
+        for (auto&& q : queries) {
+            ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
-
+func gcdValues(nums []int, queries []int64) (ans []int) {
+	mx := slices.Max(nums)
+	cnt := make([]int, mx+1)
+	cntG := make([]int, mx+1)
+	for _, x := range nums {
+		cnt[x]++
+	}
+	for i := mx; i > 0; i-- {
+		var v int
+		for j := i; j <= mx; j += i {
+			v += cnt[j]
+			cntG[i] -= cntG[j]
+		}
+		cntG[i] += v * (v - 1) / 2
+	}
+	for i := 2; i <= mx; i++ {
+		cntG[i] += cntG[i-1]
+	}
+	for _, q := range queries {
+		ans = append(ans, sort.SearchInts(cntG, int(q)+1))
+	}
+	return
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3312.Sorted GCD Pair Queries/README_EN.md

@@ -82,32 +82,145 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3312.So
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Preprocessing + Prefix Sum + Binary Search
+
+We can preprocess to obtain the occurrence count of the greatest common divisor (GCD) of all pairs in the array $\textit{nums}$, recorded in the array $\textit{cntG}$. Then, we calculate the prefix sum of the array $\textit{cntG}$. Finally, for each query, we can use binary search to find the index of the first element in the array $\textit{cntG}$ that is greater than $\textit{queries}[i]$, which is the answer.
+
+Let $\textit{mx}$ denote the maximum value in the array $\textit{nums}$, and let $\textit{cnt}$ record the occurrence count of each number in the array $\textit{nums}$. Let $\textit{cntG}[i]$ denote the number of pairs in the array $\textit{nums}$ whose GCD is equal to $i$. To calculate $\textit{cntG}[i]$, we can follow these steps:
+
+1. Calculate the occurrence count $v$ of multiples of $i$ in the array $\textit{nums}$. Then, the number of pairs formed by any two elements from these multiples must have a GCD that is a multiple of $i$, i.e., $\textit{cntG}[i]$ needs to be increased by $v \times (v - 1) / 2$;
+2. We need to exclude pairs whose GCD is a multiple of $i$ and greater than $i$. Therefore, for multiples $j$ of $i$, we need to subtract $\textit{cntG}[j]$.
+
+The above steps require us to traverse $i$ from large to small so that when calculating $\textit{cntG}[i]$, we have already calculated all $\textit{cntG}[j]$.
+
+Finally, we calculate the prefix sum of the array $\textit{cntG}$, and for each query, we can use binary search to find the index of the first element in the array $\textit{cntG}$ that is greater than $\textit{queries}[i]$, which is the answer.
+
+The time complexity is $O(n + (M + q) \times \log M)$, and the space complexity is $O(M)$. Here, $n$ and $M$ represent the length and the maximum value of the array $\textit{nums}$, respectively, and $q$ represents the number of queries.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:
+        mx = max(nums)
+        cnt = Counter(nums)
+        cnt_g = [0] * (mx + 1)
+        for i in range(mx, 0, -1):
+            v = 0
+            for j in range(i, mx + 1, i):
+                v += cnt[j]
+                cnt_g[i] -= cnt_g[j]
+            cnt_g[i] += v * (v - 1) // 2
+        s = list(accumulate(cnt_g))
+        return [bisect_right(s, q) for q in queries]
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int[] gcdValues(int[] nums, long[] queries) {
+        int mx = Arrays.stream(nums).max().getAsInt();
+        int[] cnt = new int[mx + 1];
+        long[] cntG = new long[mx + 1];
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i > 0; --i) {
+            int v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1L * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        int m = queries.length;
+        int[] ans = new int[m];
+        for (int i = 0; i < m; ++i) {
+            ans[i] = search(cntG, queries[i]);
+        }
+        return ans;
+    }
+
+    private int search(long[] nums, long x) {
+        int n = nums.length;
+        int l = 0, r = n;
+        while (l < r) {
+            int mid = l + r >> 1;
+            if (nums[mid] > x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
+        int mx = ranges::max(nums);
+        vector<int> cnt(mx + 1);
+        vector<long long> cntG(mx + 1);
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i; --i) {
+            long long v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1LL * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        vector<int> ans;
+        for (auto&& q : queries) {
+            ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
-
+func gcdValues(nums []int, queries []int64) (ans []int) {
+	mx := slices.Max(nums)
+	cnt := make([]int, mx+1)
+	cntG := make([]int, mx+1)
+	for _, x := range nums {
+		cnt[x]++
+	}
+	for i := mx; i > 0; i-- {
+		var v int
+		for j := i; j <= mx; j += i {
+			v += cnt[j]
+			cntG[i] -= cntG[j]
+		}
+		cntG[i] += v * (v - 1) / 2
+	}
+	for i := 2; i <= mx; i++ {
+		cntG[i] += cntG[i-1]
+	}
+	for _, q := range queries {
+		ans = append(ans, sort.SearchInts(cntG, int(q)+1))
+	}
+	return
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.cpp

@@ -0,0 +1,27 @@
+class Solution {
+public:
+    vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
+        int mx = ranges::max(nums);
+        vector<int> cnt(mx + 1);
+        vector<long long> cntG(mx + 1);
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i; --i) {
+            long long v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1LL * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        vector<int> ans;
+        for (auto&& q : queries) {
+            ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
+        }
+        return ans;
+    }
+};

solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.go

@@ -0,0 +1,23 @@
+func gcdValues(nums []int, queries []int64) (ans []int) {
+	mx := slices.Max(nums)
+	cnt := make([]int, mx+1)
+	cntG := make([]int, mx+1)
+	for _, x := range nums {
+		cnt[x]++
+	}
+	for i := mx; i > 0; i-- {
+		var v int
+		for j := i; j <= mx; j += i {
+			v += cnt[j]
+			cntG[i] -= cntG[j]
+		}
+		cntG[i] += v * (v - 1) / 2
+	}
+	for i := 2; i <= mx; i++ {
+		cntG[i] += cntG[i-1]
+	}
+	for _, q := range queries {
+		ans = append(ans, sort.SearchInts(cntG, int(q)+1))
+	}
+	return
+}

solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.java

@@ -0,0 +1,41 @@
+class Solution {
+    public int[] gcdValues(int[] nums, long[] queries) {
+        int mx = Arrays.stream(nums).max().getAsInt();
+        int[] cnt = new int[mx + 1];
+        long[] cntG = new long[mx + 1];
+        for (int x : nums) {
+            ++cnt[x];
+        }
+        for (int i = mx; i > 0; --i) {
+            int v = 0;
+            for (int j = i; j <= mx; j += i) {
+                v += cnt[j];
+                cntG[i] -= cntG[j];
+            }
+            cntG[i] += 1L * v * (v - 1) / 2;
+        }
+        for (int i = 2; i <= mx; ++i) {
+            cntG[i] += cntG[i - 1];
+        }
+        int m = queries.length;
+        int[] ans = new int[m];
+        for (int i = 0; i < m; ++i) {
+            ans[i] = search(cntG, queries[i]);
+        }
+        return ans;
+    }
+
+    private int search(long[] nums, long x) {
+        int n = nums.length;
+        int l = 0, r = n;
+        while (l < r) {
+            int mid = l + r >> 1;
+            if (nums[mid] > x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}