Create day002sol02.cpp

Author: Naman2251

Problems/Mathematics/Day-02/sol/Naman Pal/day002sol02.cpp

@@ -0,0 +1,65 @@
+Submission LInk: https://codeforces.com/contest/1475/submission/355652328
+
+/*
+E. Advertising Agency: We are given n bloggers and each blogger has at max n followers. We have to find the number of ways to select k 
+bloggers so that the total number of their followers is maximum possible.
+
+So first I made an array to count the number of followers of each blogger in an array and then iterated through the array from last checking if the number of bloggers 
+having that many followers are equal or greater or less than k. If they are less than k then all need to be counted in order to get the maximum so I subtracted 
+those blogger's count from k. If the number of bloggers with index number of followers is equal to or greater than k then we have to select any k of them so we 
+apply C(no.of blogger with given no. of follower, k) and then break through the loop as our work is done. 
+Now we need to handle the time complexity. For nCr we need to again and again find factorial so in order to remove that problem I made an array of factorial from 0 
+to 1000. Since array size is small therefore it take O(1) T.C. and we can get factorial of any number in O(1).Also while calculating factorial using dp I made sure 
+to take mod in each multiplication step. Now to find nCr we use the formula (n! / (r! * (n-r)!)). However, since we are working with modulo arithmetic, division is
+not straightforward. We multiply by the modular inverse instead. By Fermat's Little Theorem, the modular inverse of a number a modulo m (where m is prime) is 
+a^{m-2}. The function `binexp` calculates this power efficiently in O(log n) time. 
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+#define fastio ios::sync_with_stdio(false); cin.tie(NULL);
+ 
+const long long mod=1e9+7;
+ 
+vector<long long> fact(1001);
+ 
+long long binexp(long long a, long long b) {
+    long long ans = 1;
+    a %= mod;
+    while (b) {
+        if (b & 1) ans = ans * a % mod;
+        a = a * a % mod;
+        b >>= 1;
+    }
+    return ans%mod;
+}
+ 
+long long C(long long n, long long r) {
+    if (r < 0 || r > n) return 0;
+    return fact[n] * binexp(fact[r], mod - 2) % mod * binexp(fact[n - r], mod - 2) % mod;
+}
+ 
+int main() {
+    fastio;
+    fact[0]=1;
+    for(long long i=1; i<1001; i++) fact[i]=(fact[i-1]*i)%mod;
+    long long t;
+    cin>>t;
+    while(t--) {
+        long long n,k;
+        cin>>n>>k;
+        long long cnt[n+1]={0};
+        for(long long i=0; i<n; i++) {
+            long long a;
+            cin>>a;
+            cnt[a]++;
+        }
+        for(long long i=n; i>=0; i--) {
+            if(cnt[i]>=k) {
+                cout<<C(cnt[i], k)%mod<<endl;
+                break;
+            }
+            else k-=cnt[i];
+        }
+    }   
+}