190715-kappa-组合数学+容斥原理

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 namespace Moxing{
 9     const int N=1e4+20,lim=1e4;
10     int n,p,mod,fac[N],lim[N];
11     int mul(int a,int b){
12         int ans=0;
13         while(b){
14             if(b&1){
15                 ans=(ans+a)%mod;
16             }
17             b>>=;
18             a=(a+a)%mod;
19         }
20         return ans;
21     }
22     int power(int a,int b){
23         int ans=1;
24         while(b){
25             if(b&1){
26                 ans=mul(ans,a); 
27             }
28             b>>=1;
29             a=mul(a,a);
30         } 
31         return ans;
32     }
33     void first(){
34         fac[0]=1;
35         for(int i=1;i<=lim;i++){
36             fac[i]=mul(fac[i-1],i);
37         }
38         inv[lim]=power(fac[lim],mod-2);
39         for(int i=lim;i;i--){
40             inv[i-1]=mul(inv[i],i);
41         }
42     }
43     int c(int n,int m){
44         return mul(fac[n],mul(inv[m],inv[n-m]));
45     }
46     int f[N],g[N];
47     struct main{
48         main(){
49             scanf("%d%d",&n,&p),mod=n=10000?998244353:1e4+7;
50             first(),f[0]=g[0]=1;
51             for(int i=1;i<=n;i++){
52                 for(int j=1;j<=i;j++){
53                     if(j&1) f[i]=(f[i]+mul(power(2,mul(j,i-j)),mul(c(i,j),f[i-j])))%mod;
54                     else f[i]=(f[i]-mul(power(2,mul(j,i-j)),mul(c(i,j),f[i-j]))+mod)%mod;
55                 }
56                 g[i]=f[i];
57                 for(int j=1;j<i;j++) g[i]=(g[i]-mul(c(i-1,j-1),mul(g[j],f[i-j]))+mod)%mod;
58             }
59             printf("%d\n",p?g[n]:f[n]);
60             exit(0);
61         }
62     }UniversalLove;
63 }
64 int main(){
65     Moxing::main();
66 }

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int ans[] = {25, 18, 543, 8939, 5503, 6794, 8693, 7469, 3864, 185198037};
 5 
 6 inline int gid(int n, int p) {
 7     if(n == 3) return 1 + p;
 8     if(n == 4) return 3;
 9     if(n == 100) return 4 + p;
10     if(n == 300) return 6 + p;
11     if(n == 5000) return 8 + p;
12     return 10;
13 }
14 
15 int main() {
16     static int n, p;
17     freopen("kappa.in", "r", stdin);
18     freopen("kappa.out", "w", stdout);
19     scanf("%d%d", &n, &p), printf("%d\n", ans[gid(n, p) - 1]);
20     return 0;
21 }