Added solution for NMNMX problem

NMNMX

@@ -0,0 +1,128 @@
+#include<bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+ ll crr[5005][5005]={0};
+
+ //*****************************************************************************
+
+ ll modInv(ll a,ll m=1000000007)
+ {
+        ll m0 = m;
+        ll y = 0, x = 1;
+        if (m == 1)
+                return 0;
+        while (a > 1){
+                // q is quotient
+                ll q = a / m;
+                ll t = m;
+                // m is remainder now, process same as
+                // Euclid's algo
+                m = a % m, a = t;
+                t = y;
+                // Update y and x
+                y = x - q * y;
+                x = t;
+        }
+        // Make x positive
+        if (x < 0)
+    {
+       x += m0;
+    }
+        return x;
+}
+
+ll modDivide(ll a,ll b,ll m=1000000007)
+{
+       // a = a % m;
+        ll inverse= modInv(b,m);
+        return (inverse*a)%m;
+}
+
+
+//******************************************************************************
+
+ll modex(ll x,ll y,ll lebhai)
+{
+    if(y<=0)
+        return 1;
+    else if(y%2==0)
+        return modex((x%lebhai*x%lebhai)%lebhai,y/2,lebhai);
+    else
+        return (x%lebhai*(modex((x%lebhai*x%lebhai)%lebhai,(y-1)/2,lebhai))%lebhai)%lebhai;
+}
+int main()
+{ ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+ll t,p=1000000006;
+
+
+for (ll i = 0; i <= 5000; i++)
+    {
+        for (ll j= 0; j <= i; j++)
+
+         {
+
+            if (i == j||j == 0  )
+                {crr[i][j] = 1;
+                }
+
+            else
+                 {crr[i][j] =  (crr[i-1][j-1] + crr[i-1][j])%p;
+    }}
+
+}
+         cin>>t;
+         while(t--)
+         {
+
+         ll n,k,it,in,pr=1,p3,p4,p2;
+         cin>>n>>k;
+         ll a[n];
+         for(it=0;it<n;it++)
+         {
+             cin>>a[it];
+         }
+         sort(a,a+n);
+         // if(n%2!=0)
+     //  {
+        for( in=1;in<n-1;in++)
+        { p2=(n-in-1)>=(k-1)? crr[n-in-1][k-1]:0;
+         p3=(in>=(k-1))?crr[in][k-1]:0;
+        p4=((n-1)>=(k-1))?crr[n-1][k-1]:0;
+
+         ll ans1=modex(a[in],p4%p,1000000007);
+           ll ans2=modex(a[in],((p2+p3)%p),1000000007);
+          pr= (pr%1000000007*(modDivide(ans1,ans2,1000000007))%1000000007)%1000000007;
+          // b[in]=(p4- p2-p3)%1000000006;
+           // b[n-in-1]=b[in];
+        }
+      // }
+  /*else
+       {
+
+            for( in=1;in<=(n/2)-1;in++)
+           { p2=(n-in-1)>=(k-1)? crr[n-in-1][k-1]:0;
+         p3=(in>=(k-1))?crr[in][k-1]:0;
+         p4=((n-1)>=(k-1))?crr[n-1][k-1]:0;
+          // b[in]=(p4- p2-p3)%1000000006;
+          //  b[n-in-1]=b[in];
+        }
+
+       }*/
+  /*for( in=1;in<=n-2;in++)
+            {cout<<b[in]<<" ";
+        }
+cout<<"hello\n"<<endl;*/
+        /*  for(in=1;in<n-1;in++)
+       {
+           ll ans1=modex(a[in],p4,1000000007);
+           ll ans2=modex(a[in],((p2+p3)%p),1000000007);
+          pr= (pr%1000000007*(modDivide(ans1,ans2,1000000007))%1000000007)%1000000007;
+           //(pr%1000000007*(modex(a[in],b[in],1000000007))%1000000007)%1000000007;
+       }*/
+       cout<<pr<<endl;
+
+    }
+    return 0;
+
+}