2019icpc Shenyang network game

J title:   

Meaning of the questions: to give you a view of uncertain size n, and this figure uncertain degree and out of each point is 1. Then a feature of this map is certainly composed by a number of Euler. Can get recursive formula is n!

Then you look at it there is a feature. It requires that each point must take no more than x can return to their own sides. . . . . . . (N <= 2 * x <2e9)  

We certainly can only think from the negative side, it does not meet the conditions are removed as (n <= 2 * x <2e9) This condition may be doomed to think so. . . . . . .

Because p (m) = c (m, n) * (m-1)! * (Nm)! / N! = 1 / m so for each probability is 1 / i (x <i <= n)

Pretreatment what you can, Here is the code:

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <bitset>
 5 #include <iostream>
 6 #include <algorithm>
 7 typedef long long ll;
 8 using namespace std;
 9 const ll mod=(ll)(1e9+7);
10 const int maxn=(int)(1e6+10);
11 ll sum[maxn];
12 int t;
13 ll x,n;
14 
15 ll fast_pow(ll a,ll b){
16    ll res=1;
17    while(b){
18        if(b%2==1) res=(res*a)%mod;
19        a=(a*a)%mod;
20        b=b/2;
21    }
22    return res;
23 }
24 
25 void init(){
26     sum[0]=0;
27     for(ll i=1;i<maxn;i++){
28         sum[i]=(sum[i-1]+fast_pow(i,mod-2))%mod;
29     }
30 }
31 
32 
33 
34 int main(){
35     init();
36     scanf("%d",&t);
37     while(t--){
38         scanf("%lld%lld",&n,&x);
39         if(x>=n){
40             printf("1\n");
41             continue;
42         }else{
43             ll ans1=1,ans2=1;
44             for(ll i=x+1;i<=n;i++)  ans1=(ans1*i)%mod;
45             ans2=(ans1-(ans1*(sum[n]-sum[x]+2*mod))%mod+2*mod)%mod;
46             //cout<<ans1<<"  "<<ans2<<endl;
47             printf("%lld\n",ans2*fast_pow(ans1,mod-2)%mod);
48         }
49     }
50     return 0;
51 }

View Code