Merge pull request #503 from Aiyaan-Mahajan/day8-q2

Day 8 Q2

Author: Dheeraj-06

Problems/Mathematics/Day-08/Aiyaan_Mahajan/q2sol.cpp

@@ -0,0 +1,78 @@
+/*
+DAY 8
+Q2 F. Expected Median
+
+*/
+/*
+
+
+For an odd-length subsequence, its median is **1** iff it has at least ((k+1)/2) ones (because the array is binary).
+So the answer equals the number of subsequences of length (k) that contain at least ((k+1)/2) ones.
+Count subsequences with exactly (t) ones using (\binom{c1}{t}\binom{c0}{k-t}) and sum over all (t \ge (k+1)/2), modulo (10^9+7).
+
+
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+const long long MOD = 1'000'000'007;
+const int MAXN = 200000 * 2 + 5;
+
+long long fact[MAXN], invfact[MAXN];
+
+long long power(long long a, long long b) {
+    long long r = 1;
+    while (b) {
+        if (b & 1) r = r * a % MOD;
+        a = a * a % MOD;
+        b >>= 1;
+    }
+    return r;
+}
+
+long long nCr(int n, int r) {
+    if (r < 0 || r > n) return 0;
+    return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD;
+}
+
+int main() {
+    
+
+    
+    fact[0] = 1;
+    for (int i = 1; i < MAXN; i++) fact[i] = fact[i - 1] * i % MOD;
+    invfact[0] = 1;
+    invfact[MAXN - 1] = power(fact[MAXN - 1], MOD - 2);
+    for (int i = MAXN - 2; i >= 1; i--)
+        invfact[i] = invfact[i + 1] * (i + 1) % MOD;
+
+    int t;
+    cin >> t;
+    while (t--) {
+        int n, k;
+        cin >> n >> k;
+
+        int c1 = 0, c0 = 0;
+        for (int i = 0; i < n; i++) {
+            int x; cin >> x;
+            if (x) c1++;
+            else c0++;
+        }
+
+        int m = (k + 1) / 2;
+        long long ans = 0;
+
+        for (int t = m; t <= k && t <= c1; t++) {
+            ans = (ans + nCr(c1, t) * nCr(c0, k - t)) % MOD;
+        }
+
+        cout << ans % MOD << "\n";
+    }
+
+    return 0;
+}
+
+//Tc =O(n+k) per test  SC = O(n)
+/*
+My submission : https://codeforces.com/contest/1999/submission/356297537
+*/