Solve CCC and CCO problems

CCC/ccc04s4.cpp

@@ -0,0 +1,44 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main()
+{
+    long long sx, sy, sz, tx, ty, tz;
+    int d;
+    char t;
+    scanf("%lli%lli%lli%lli%lli%lli", &sx, &sy, &sz, &tx, &ty, &tz);
+    tx -= sx;
+    ty -= sy;
+    tz -= sz;
+    long long ans = tx * tx + ty * ty + tz * tz;
+    while (1)
+    {
+        scanf("%i %c", &d, &t);
+        if ((tx - d < 0 && tx > 0) || (tx - d > 0 && tx < 0))
+            ans = min(ans, ty * ty + tz * tz);
+        tx -= d;
+        ans = min(ans, tx * tx + ty * ty + tz * tz);
+        if (t == 'E')
+            return 0 * printf("%.2lf\n", sqrt(ans));
+        switch(t)
+        {
+            case 'R':
+                swap(tx, ty);
+                tx = -tx;
+            break;
+            case 'L':
+                swap(tx, ty);
+                ty = -ty;
+            break;
+            case 'U':
+                swap(tx, tz);
+                tz = -tz;
+            break;
+            case 'D':
+                swap(tx, tz);
+                tx = -tx;
+            break;
+        }
+    }
+}

CCO/cco16p2.cpp

@@ -0,0 +1,74 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const long double PI = acos(-1);
+
+struct point
+{
+    int x, y;
+};
+
+struct bunny : point
+{
+    int w;
+};
+
+static inline long double ang(point a, point o)
+{
+    int yy = a.y-o.y;
+    int xx = a.x-o.x;
+    int g = abs(__gcd(xx, yy));
+    return atan2(yy/g, xx/g);
+}
+
+static inline bool eq(long double a, long double b)
+{
+    return abs(a-b) < 1e-9;
+}
+
+int main()
+{
+    int n;
+    scanf("%i", &n);
+    vector<bunny> arr(n);
+    for (auto &x : arr)
+        scanf("%i%i%i", &x.x, &x.y, &x.w);
+    int ans = 0;
+    for (int z = 0; z < n; z++)
+    {
+        map<long double, int> mp;
+        int sum = 0;
+        for (int y = 0; y < n; y++)
+        {
+            if (z == y)
+                continue;
+            sum += arr[y].w;
+            mp[ang(arr[y], arr[z])] += arr[y].w;
+        }
+        vector<pair<long double,int>> cur(mp.size()*2);
+        int cc = 0;
+        for (auto &x : mp)
+            cur[cc++] = x;
+        for (auto &x : mp)
+        {
+            cur[cc] = x;
+            cur[cc].first += 2*PI;
+            cc++;
+        }
+        int sz = mp.size();
+        int curs = 0;
+        for (int x = 0, y = 0; x < sz; x++)
+        {
+            long double VV = cur[x].first + PI;
+            while (y < cur.size() && cur[y].first < VV)
+            {
+                curs += cur[y].second;
+                y++;
+            }
+            ans = max(ans, max(sum-curs + arr[z].w, arr[z].w + curs + eq(cur[y].first, VV)*cur[y].second));
+            curs -= cur[x].second;
+        }
+    }
+    printf("%i\n", ans);
+}