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Other_Tasks/CF1971_F_Circle_Perimeter.cpp

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+/**CF1971F
+ * Finding the number of lattice points (x, y) suc that r^2 <= x^2 + y^2 < (r + 1)^2
+ * We can restrict ourselves to finding (x, y) with x >= 1 and y >= 0 then multiplying by 4 due to
+ * rotational symmetry.
+ *
+ * For each 1 <= x <= r, we are interested in finding the number of y such thar r^2 - x^2 <= y^2 <
+ * (r + 1)^2 - x^2 To do this, we can binary search our precomputed array of square numbers twice.
+ * Might be non-trivial to reason about how exactly to do this but just do it.
+ * 
+ * Time: O(r log r), Space: O(r)
+ */
+#pragma GCC optimize("Ofast")
+#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
+#pragma GCC optimize("unroll-loops")
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long ll;
+
+ll r, rs, rp1s;
+vector<ll> squares;
+int main()
+{
+    ll rmax = 1e5 + 1;
+    squares.resize(rmax + 2);
+    for (ll i = 1; i <= (rmax + 1); i++) {
+        squares[i] = i * i;
+    }
+
+    int tc;
+    cin >> tc;
+    while (tc--) {
+        cin >> r;
+        rs = squares[r];
+        rp1s = squares[r + 1];
+
+        ll ans = 0;
+        for (int x = 1; x <= r; x++) {
+            // find the largest i such that squares[i] < rp1s - squares[x]
+            ll bound = rp1s - squares[x];
+            ll high = lower_bound(squares.begin(), squares.end(), bound) - squares.begin() - 1;
+
+            // find the largest i such that squares[i] < rs - squares[x]
+            bound = rs - squares[x];
+            ll low = lower_bound(squares.begin(), squares.end(), bound) - squares.begin() - 1;
+
+            ans += high - low;
+        }
+        cout << ans * 4LL << endl;
+    }
+    return 0;
+}