I was doing a question Advertising Agency Your text to link here... where i need to find NcR but if N is very large i had to use modulo of 10^9+7 but for some reason my answer gets wrong on test 6 where n is very big. I think the problem is not using modulo properly but i cant figure it out. 1475E][SUBMISSION:121841381 - Advertising Agency any help is appreciated. In my code n=to and r=bef;
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
//#include <chrono>
//using namespace std::chrono;
long long multi(long long a, long long b, long long mod){
return (a * b) % mod;
}
long long power(long long a, long long b, long long mod){
long long powans = 1;
for(; b > 0; a = multi(a, a, mod), b /= 2) if(b % 2 == 1) powans = multi(powans, a, mod);
return powans;
}
void fastscan(int &x)
{
bool neg=false;
register int c;
x =0;
c=getchar();
if(c=='-')
{
neg = true;
c=getchar();
}
for(;(c>47 && c<58);c=getchar())
x = (x<<1) + (x<<3) +c -48;
if(neg)
x *=-1;
}
int gcd(long long int a,long long int b){
if(a%b==0)
return b;
else
return gcd(b,a%b);
}
long long int fact(long long int to,long long int bef){
long long sol=1;
if(to-bef<bef)
bef=to-bef;
long long int p=bef;
long long int n=1,d=1;
while(p--){
n=n*to;
d=d*bef;
long long int m=gcd(n,d);
n=n/m;
d=d/m;
n=n%1000000007;
d=d%1000000007;
to--;
bef--;
}
return n;
}
void solve(){
int n,k;
cin>>n>>k;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
sort(arr,arr+n,greater<int>());
int value=arr[k-1];
long long int bef=0,to=0;
for(int i=0;i<n;i++){
if(arr[i]==value){
to++;
if(i<=k-1)
bef++;
}
}
long long int sol=1;
sol=fact(to,bef);
cout<<sol%1000000007<<endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
freopen("error.txt","w",stderr);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
//auto start = high_resolution_clock::now();
int t;
cin>>t;
while(t--) {
solve();
}
// auto stop = high_resolution_clock::now();
// auto duration = duration_cast<microseconds>(stop - start);
// cout << duration.count() << endl;
return 0;
}