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Author: capricemoto

Problems/Mathematics/Day-08/soln/Samarth/sol2.cpp

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+#include <bits/stdc++.h>
+#define endl '\n'
+typedef long long ll;
+
+using namespace std;
+
+/*
+PROBLEM STATMENT:
+Arul has a binary array∗ a of length n
+
+.
+
+He will take all subsequences†
+of length k (k is odd) of this array and find their median.‡
+
+What is the sum of all these values?
+
+As this sum can be very large, output it modulo 109+7
+. In other words, print the remainder of this sum when divided by 109+7.
+
+APPROACH:
+to get the ans we just needed the different combinations of sequence where the median would be one;
+our ans was nothing but diff combinations so for median one (k+1)/2th element must be one so that means there must atleast (k+1)/2 ones 
+and atmost k ones now then we have to calculte no of ways of selecting ones and zero from given ones;
+
+now as we would have to compute ncr many times we used precomputation of fac,and inv fac
+to trade off memory for speed.
+
+TIME COMPLEXITY:O(MAXN);
+SPACE COMPLEXITY:O(MAXN);
+
+SUBMISSION LINK:
+https://codeforces.com/contest/1999/submission/356201034
+
+*/
+
+const ll h=1e9+7;
+const ll maxn=2*(1e5)+1;
+
+ll fac[maxn],inv[maxn];
+
+ll binexp(int a,int b){
+	if(b==0)return 1;
+	
+	ll set=binexp(a,b/2);
+	
+	set=(set*set)%h;
+	if (b&1)
+		return (set*a)%h;
+	else 
+		return set;
+
+}
+
+
+void precomp(){
+	fac[0]=1;
+	for(int i=1;i<maxn;i++)fac[i]=(fac[i-1]*i)%h;
+	inv[maxn-1]=binexp(fac[maxn-1],h-2);
+	for(int i=maxn-2;i>=0;i--)inv[i]=(inv[i+1]*(i+1))%h;
+}
+
+
+
+long long comb(int s,int t){
+	ll a= fac[s];
+	ll b=inv[s-t];
+	ll c=inv[t];
+	return (((a*b)%h)*c)%h;
+}
+
+
+
+
+void solve(){
+	int n,k;cin>>n>>k;int y;int x=0;ll ans=0;
+	for(int i=0;i<n;i++){
+		cin>>y;
+		if(y==1)x++;
+	}
+
+
+	for(int i=max((k+1)/2,k-n+x);i<=min(x,k);i++)
+		ans=(ans +(comb(x,i)*comb(n-x,k-i))%h)%h;
+
+	cout<<ans<<endl;
+}
+
+int main(){
+    ios::sync_with_stdio(0); cin.tie(0);
+	int t;cin>>t;
+	precomp();
+	while(t--)
+		solve();
+	
+}