Added Tutorial for LOj-1161 Extreme GCD (en)

1161/en.md

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+# LOJ 1161 - Extreme GCD
+
+## Summary
+The problem is asking for the number of ways to choose four positive integers, which may or may not be distinct, from a given set of N positive integers, such that their greatest common divisor (GCD) is 1.
+
+## Prerequisite
+i) Basic Permutation and Combination. 
+
+ii) Finding divisors of an integer N in O( $\sqrt{N}$ ) complexity.
+
+## Solution
+Let's generalize the problem a bit. We define:
+
+**div_cnt[i]** = how many of those N numbers have a divisor equal to **i**.
+
+**ans[k]** = number of ways choosing **4** integers from the set of N numbers such that their gcd is **equal to k**.
+
+Now in order for having gcd equal to **k** those 4 numbers must have **k** as their common divisor but not a multiple of **k** (greater than **k**) as in that case gcd will not be equal to **k** (will be a multiple of **k** greater than **k** itself). If we set **ans[k]** = **nC4**( **div_cnt[k]** ) then we would have wronged ourselvs, because **ans[k]** will going to count some selections where gcd of those 4 numbers is a multiple of **k** but not **k**. So in this scenario we have to substract all **ans[k*i]** (i > 1) from **ans[k]** so that the **ans[k]** will be the required answer. Having discussed the solution we are bound to calculate the **ans** array in decreasing order as every **ans[i]** depends on the multiples of **i**.
+
+## Complexity
+- Time Complexity: O(T * N * $log{_2}{N}$ ).
+- Memory Complexity: O(N).
+
+## Code
+
+### C++
+
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+
+typedef long long ll;
+
+
+const int MAXN = 1e4;
+
+inline ll nC4(ll n) {
+    return (n*(n-1)*(n-2)*(n-3)) / 24;
+}
+
+int main(int argc, const char *argv[]) {
+
+    // for fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    for(int ts = 1; ts <= t; ++ts) {
+		int n;
+        cin >> n;
+
+        vector <int> div_cnt(MAXN+1);
+        for(int i = 0; i < n; ++i) {
+            int x;
+            cin >> x;
+
+            for(int j = 1; j*j <= x; ++j) {
+                if (x % j) continue;
+                div_cnt[j]++;
+                if (j*j != x) div_cnt[x/j]++;
+            }
+        }
+
+        vector <ll> ans(MAXN+1);
+        for(int i = MAXN; i >= 1; --i) {
+            ans[i] = nC4(div_cnt[i]);
+            for(int j = i+i; j <= MAXN; j += i) {
+                ans[i] -= ans[j];
+            }
+        }
+
+        cout << "Case " << ts << ": " << ans[1] << '\n';
+	}
+
+    return 0;
+} 
+```