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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. typedef long long ll;
  5. const int mod=1000000007;
  6. vector<ll> fact(1e5+1, 1);
  7.  
  8. ll modInv(ll a, ll e){
  9.     ll res=1;
  10.     while(e>0){
  11.         if (e%2){
  12.             res=(res*a)%mod;
  13.             e--;
  14.         }
  15.         else{
  16.             a=(a*a)%mod;
  17.             e/=2;
  18.         }
  19.     }
  20.     return res;
  21. }
  22. ll nCr(ll n, ll r){
  23.     ll invR = modInv(fact[r], mod-2);       
  24.     ll invNR = modInv(fact[n-r], mod-2);  
  25.     ll denom = (invR * invNR) % mod;     
  26.     return (fact[n] * denom) % mod;
  27. }
  28. int main() {
  29.     for(int i=1; i<=1e5; i++) fact[i]=(fact[i-1]*i)%mod;
  30. 	int t; cin>>t;
  31. 	while(t--){
  32. 	    ll n, k; cin>>n>>k;
  33. 	    ll p=0LL;
  34. 	    for(int i=1; i<=n; i++){
  35. 	        int nxt=k-i;
  36. 	        if (nxt>=1 && nxt<i) p++;
  37. 	    }
  38. 	    ll trail=1, res=0LL;
  39. 	    for(ll i=0; i<p; i++){
  40. 	        ll d = (i % 2) ? -1 : 1;
  41.             trail = (trail * 2) % mod; 
  42.             ll ways = nCr(p, i + 1);
  43.             ll term = (ways * trail) % mod;
  44.             term = (term * fact[n - i - 1]) % mod; 
  45.             term = (d * term + mod) % mod;
  46.             res = (res + term) % mod; 
  47. 	    }
  48. 	    cout << res << endl;
  49. 	}
  50.  
  51. }
  52.  
Success #stdin #stdout 0.01s 5284KB
comments (?)
stdin
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10
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8 10
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stdout
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4
48
1440
22560
80640
725760
0
422027771
488057398
656420449
https://ideone.com/mMM7wK
Permutation Pair:
You're given two integers N and K. Count the number of permutations P of the integers 1 to N satisfy the following condition: There exists at least one index i ( 1 ≤ i < N ) such that Pi + Pi+1 = K. Since the count can be very large, print it modulo 10^9 + 7.
language:
C++14 (gcc 8.3)
created:
15 days ago
visibility:
public

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