Educational Codeforces Round 71 Editorial D

It is a permutation of the subject.

Observed up to a total of N! Permutations embodiment, the first keyword, if there are the same elements a1, a2 ... so bad permutation C1 = a1 * a2 * ... identical to the second key, obtains C2 . However, if the total of minus two kinds of bad permutation is clearly incorrect, because it is possible to delete the duplicate arrangement, then put this together like a repeat, is the first keyword orderly, also ordered the second keyword . The final ans = N! -C1-C2 + C3.

Write your own time should be in modulo bombed, and found a very good writing. for reference

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn=3e5+5;
 5 const ll mod=998244353;
 6 
 7 pair <int,int > a[maxn];
 8 ll d[maxn];
 9 ll fi[maxn],se[maxn];
10 map<pair<int,int>,int> M;
11 int n;
12 ll sum=1;ll e=1,b=1,c=1;
13 int main(){
14     cin>>n;
15     d[0]=1;
16     for(int i=1;i<=n;i++){
17         cin>>a[i].first>>a[i].second;
18         d[i]=(d[i-1]*i)%mod;
19         e=e*(++fi[a[i].first])%mod;
20         b=b*(++se[a[i].second])%mod;
21         c=c*(++M[a[i]])%mod;
22     }
23     sort(a+1,a+n+1);
24     
25     int now=a[1].second;int sign=1;
26     for(int i=2;i<=n;i++){
27         if(a[i].second>=now) now=a[i].second;
28             else 
29             {
30                 sign=0;
31                 break;
32             }
33     }
34     ll ans;
35     if(sign) 
36     {
37         ans=(d[n]+c+2*mod-e-b)%mod;
38     }
39     else ans=(d[n]+2*mod-e-b)%mod;
40     cout <<ans<<"\n";
41     
42     
43     return 0;
44 }