Add Tutorial for LOJ-1077 How Many Points? (en)

1077/en.md

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+# LOJ 1077 - How Many Points?
+
+
+## Solution
+The number of lattice points lying on the segment **A(x1, y1) -> B(x2, y2)** is same for segment **A(0, 0) -> B(x2 - x1, y2 - y1)** because co-ordinate translation doesn't change the relative distance of lattice points from each other.
+The number of lattice points will also be equal to that of segment **A(0, 0) -> B(|x2 - x1|, |y2 - y1|)** since sign (+/-) only tells us in which quadrants the segment will fall.
+Number of lattice points is length dependent not quadrant dependent.
+
+So, how to calculate the number of lattice points on any segment **A(0, 0) -> B(x, y)** where **x, y >= 0**? The answer is: **gcd(x, y) + 1**. Why?
+
+Suppose, **g = gcd(x, y)** then, all the lattice points are:
+
+**(0 * x/g, 0 * y/g), (1 * x/g, 1 * y/g), . . . . . . . . . . . . . . . , ((g-1) * x/g, (g-1) * y/g), (g * x/g, g * y/g)** total of **(g + 1)** points with integer abscissas and ordinates.
+
+But what's the proof they lie on the segment **AB** and there can't be any other lattice points? 
+
+See: https://math.stackexchange.com/questions/628117/how-to-count-lattice-points-on-a-line
+
+## Complexity
+- Time Complexity: O(T * lg(N)). Where N = **max(a, b)** of **gcd(a, b)**. [Check](https://stackoverflow.com/questions/3980416/time-complexity-of-euclids-algorithm) for the time complexity of Euclid's GCD algorithm.
+- Memory Complexity: O(1).
+
+## Code
+
+### C++
+
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+
+    // For fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    for(int ts = 1; ts <= t; ++ts) {
+        pair <long long, long long> A, B;
+        cin >> A.first >> A.second >> B.first >> B.second;
+
+        cout << "Case " << ts << ": " << __gcd(abs(A.first - B.first), abs(A.second - B.second)) + 1 << '\n';
+    }
+
+    return 0;
+}
+```