Merge pull request #199 from ishantripathi64/main

solved second question of day 2

Author: zonedoutdg

Problems/Mathematics/Day-02/sol/Ishantripathi/solution2.cpp

@@ -0,0 +1,133 @@
+#include <bits/stdc++.h>
+
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+
+using namespace std;
+using namespace __gnu_pbds;
+
+typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key
+// less makes it sorted greater makes it reverse sorted less equal makes it like multiset
+// we can also change the data type of the pbds
+
+// Fast I/O
+#define FAST_IO                  \
+    ios::sync_with_stdio(false); \
+    cin.tie(NULL);
+#define pb push_back
+#define in insert
+#define int long long
+#define all(x) (x).begin(), (x).end()
+
+// Typedefs for convenience
+typedef vector<int> vi;
+typedef pair<int, int> pii;
+
+// Constants
+const int INF = 1e18;
+const int MOD = 1e9 + 7;
+
+// Debug (can be disabled in contests)
+#ifdef DEBUG
+#define dbg(x) cerr << #x << " = " << x << '\n';
+#else
+#define dbg(x)
+#endif
+
+// gcd
+int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
+
+// lcm
+int lcm(int a, int b)
+{
+    if (a == 0 || b == 0)
+        return 0;
+    return abs(a * b) / gcd(a, b);
+}
+
+// Binary exponentiation (modular)
+long long binexp(long long a, long long b, long long m)
+{
+    a %= m;
+    long long res = 1;
+    while (b > 0)
+    {
+        if (b & 1)
+            res = res * a % m;
+        a = a * a % m;
+        b >>= 1;
+    }
+    return res;
+}
+vi fac(1001);
+
+void solve()
+{
+    int n, k;
+    cin >> n >> k;
+    vi v(n + 1, 0);
+    for (int i = 1; i <= n; i++)
+    {
+        cin >> v[i];
+    }
+
+    sort(all(v));
+    int j = n - k + 1;
+    int ele = v[n - k + 1];
+    map<int, int> mp;
+    for (int i = 1; i <= n; i++)
+    {
+        mp[v[i]]++;
+    }
+
+    int i;
+    for (i = j; i <= n; i++)
+    {
+        if (v[i] != ele)
+        {
+            break;
+        }
+    }
+    int left = i - j;
+    int p = fac[mp[ele]];
+    // cout<<p<<endl;
+    int q = fac[left];
+    int s = fac[mp[ele] - left];
+    q = binexp(q, MOD - 2, MOD);
+    s = binexp(s, MOD - 2, MOD);
+    int res = (((p % MOD) * (q % MOD)) % MOD * (s % MOD)) % MOD;
+    cout << res << endl;
+}
+
+signed main()
+{
+    FAST_IO
+    fac[0] = 1;
+    for (int i = 1; i <= 1000; i++)
+    {
+        fac[i] = (fac[i - 1] * i) % MOD;
+    }
+
+    int t = 1;
+    cin >> t;
+    while (t--)
+        solve();
+    return 0;
+}
+/*
+    So basically we're asked to count number of ways we can take k elements such that there sum is maximum now
+    if all elements were unique we could have just taken last k elements and answer would be 1
+    but since they are unique
+    we can have something like n= 6 ;k=5; 1 1 1 2 2 2 where we can choose any two of the three ones hence
+    we need to take how many of those last elments we are taking as left and total now many elements of last are there are ele
+    hence the answer would be mp[ele]Cleft which is easily solvable by using fermet's little theorem to find modulo inverse
+
+
+    time complexity o(1000)
+    space complexity o(1000)
+
+    submission id :https://codeforces.com/contest/1475/submission/355268225
+
+
+
+*/