Aqib12

dsu and ncr

Apr 19th, 2021
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  1. struct DSU {
  2.   struct Node {
  3.     int p, sz;
  4.   };
  5.   vector<Node> dsu;
  6.   int cc;
  7.  
  8.   Node &operator[](int id) { return dsu[root(id)]; }
  9.  
  10.   DSU(int n) {
  11.     dsu = vector<Node>(n);
  12.     for (int i = 0; i < n; i++) {
  13.       cc = n;
  14.       dsu[i].p = i;
  15.       dsu[i].sz = 1;
  16.     }
  17.   }
  18.   //gives the parent or root of the node
  19.   inline int root(int u) { return (dsu[u].p == u) ? u : dsu[u].p = root(dsu[u].p); }
  20.  
  21.   inline bool sameSet(int u, int v) { return root(u) == root(v); }
  22.  
  23.   void merge(int u, int v) {
  24.     u = root(u);
  25.     v = root(v);
  26.     if (u == v) return;
  27.     if (dsu[u].sz < dsu[v].sz) swap(u, v);
  28.     dsu[v].p = u;
  29.     dsu[u].sz += dsu[v].sz;
  30.     cc--;
  31.   }
  32.   // size of the set where the node belongs
  33.   inline int get_set_size(int u) { return dsu[root(u)].sz; }
  34. };
  35.  
  36.  
  37. struct BinomialCoefficient {
  38.   int MAX_N = 1e6 + 1;
  39.   const int MOD = 1e9 + 7;
  40.  
  41.   long long power(long long a, long long b, long long m) {
  42.     long long res = 1;
  43.     while (b) {
  44.       if (b & 1) res = res * a % m;
  45.       a = a * a % m;
  46.       b >>= 1;
  47.     }
  48.     return res;
  49.   }
  50.  
  51.   vector<long long> fact, inv;
  52.   vector<double> log_fact;
  53.  
  54.   BinomialCoefficient() {
  55.     fact.resize(MAX_N);
  56.     inv.resize(MAX_N);
  57.     log_fact.resize(MAX_N);
  58.   }
  59.  
  60.   void precompute() {
  61.     fact[0] = inv[0] = 1;
  62.     for (int i = 1; i < MAX_N; i++) {
  63.       fact[i] = fact[i - 1] * i % MOD;
  64.       inv[i] = power(fact[i], MOD - 2, MOD); // Fermat's little theorem
  65.     }
  66.   }
  67.  
  68.   long long nCk(int n, int k) {
  69.     if (k < 0 || k > n) return 0;
  70.     return fact[n] * inv[k] % MOD * inv[n - k] % MOD;
  71.  
  72.     // if there are only a few queries, then don't need to precompute inv[] => faster
  73.     // return fact[n] * power(fact[k], MOD - 2, MOD) % MOD * power(fact[n - k], MOD - 2, MOD) % MOD;
  74.   }
  75.  
  76. // A trick to calculate large factorial without overflowing is to take log at every step when precompute and take exponential when calculating
  77. // Don't need inv[] now because it is the same as negative log of fact
  78.   void precompute_log() {
  79.     log_fact[0] = 0.0;
  80.     for (int i = 1; i < MAX_N; i++)
  81.       log_fact[i] = log_fact[i - 1] + log(i);
  82.   }
  83.  
  84.   long long log_nCk(int n, int k) {
  85.     if (k < 0 || k > n) return 0;
  86.     return exp(log_fact[n] - log_fact[n - k] - log_fact[k]);
  87.   }
  88. };
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