[递推 || 容斥 FFT] SRM 717 div1 DerangementsStrikeBack

首先像我这种无脑的人可以大力上fft

fin!=∑j=0i(−1)j(ij)(n+i−j)!n!

然而考虑经典错排的递推公式

dn=(n−1)(dn−1+dn−2)

这个东西的递推式是 把第n个和n−1的错排其中一个交换 或 把第n个跟n−1的排列中其中一个不错排的那位交换下

这题因为最后除了 n! 那么实际上后面 n 个球都是相同的，那么考虑在第 i 位插入 i 那么要和错排好的交换或者和前 i−1 个中唯一的一个交换

那么应该是这么递推的

f[0]=1; f[1]=_n;
for (int i=2;i<=_m;i++)
f[i]=((_n+i-1)*f[i-1]+(i-1)*f[i-2])%P;

// BEGIN CUT HERE
#include<conio.h>
#include<sstream>
// END CUT HERE
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<string>
#include<set>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;

const int P=1e9+7;;
const int M[]={998244353,1004535809,469762049};
const int G[]={3,3,3};
const ll _M=(ll)M[0]*M[1];

inline ll Pow(ll a,int b,int p){
ll ret=1;
for (;b;b>>=1,a=a*a%p)
if (b&1)
ret=ret*a%p;
return ret;
}
inline ll mul(ll a,ll b,ll p){
a%=p; b%=p;
return ((a*b-(ll)((ll)((long double)a/p*b+1e-3)*p))%p+p)%p;
}

const int m1=M[0],m2=M[1],m3=M[2];
const int inv1=Pow(m1%m2,m2-2,m2),inv2=Pow(m2%m1,m1-2,m1),inv12=Pow(_M%m3,m3-2,m3);
inline int CRT(int a1,int a2,int a3){
ll A=(mul((ll)a1*m2%_M,inv2,_M)+mul((ll)a2*m1%_M,inv1,_M))%_M;
ll k=((ll)a3+m3-A%m3)*inv12%m3;
return (k*(_M%P)+A)%P;
}

const int N=264000;

struct NTT{
int P,G;
int num,w[2]
;
int R
;
void Pre(int _P,int _G,int m){
num=m; P=_P; G=_G;
int g=Pow(G,(P-1)/num,P);
w[1][0]=1; for (int i=1;i<num;i++) w[1][i]=(ll)w[1][i-1]*g%P;
w[0][0]=1; for (int i=1;i<num;i++) w[0][i]=w[1][num-i];
int L=0; while (m>>=1) L++;
for (int i=1;i<=num;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
}
void FFT(int *a,int n,int r){
for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);
for (int i=1;i<n;i<<=1)
for (int j=0;j<n;j+=(i<<1))
for (int k=0;k<i;k++){
int x=a[j+k],y=(ll)a[j+i+k]*w[r][num/(i<<1)*k]%P;
a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;
}
if (!r) for (int i=0,inv=Pow(n,P-2,P);i<n;i++) a[i]=(ll)a[i]*inv%P;
}
}ntt[3];

int n,m;
int a[3]
;
int A
,B
,C
,D
;

ll fac
,inv
;

ll f
;

class DerangementsStrikeBack{
public:
int count(int _n, int _m){
/*fac[0]=1; for (int i=1;i<=_m;i++) fac[i]=fac[i-1]*i%P;
inv[1]=1; for (int i=2;i<=_m;i++) inv[i]=(ll)(P-P/i)*inv[P%i]%P;
inv[0]=1; for (int i=1;i<=_m;i++) inv[i]=inv[i-1]*inv[i]%P;
for (int i=0;i<=_m;i++) A[i]=(i&1)?(P-inv[i])%P:inv[i];
ll tmp=1;
for (int i=0;i<=_m;i++,tmp=tmp*(_n+i)%P)
B[i]=inv[i]*tmp%P;

for (m=1;m<=2*_m;m<<=1);
for (int i=0;i<3;i++) ntt[i].Pre(M[i],G[i],m);
for (int i=0;i<3;i++){
memcpy(C,A,sizeof(int)*(m+5)); memcpy(D,B,sizeof(int)*(m+5));
ntt[i].FFT(C,m,1); ntt[i].FFT(D,m,1);
for (int j=0;j<m;j++) C[j]=(ll)C[j]*D[j]%ntt[i].P;
ntt[i].FFT(C,m,0);
for (int j=0;j<m;j++) a[i][j]=C[j];
}
int ans=0;
for (int i=1;i<=_m;i++)
ans^=(ll)CRT(a[0][i],a[1][i],a[2][i])*fac[i]%P;
return ans;*/
f[0]=1; f[1]=_n;
for (int i=2;i<=_m;i++)
f[i]=((_n+i-1)*f[i-1]+(i-1)*f[i-2])%P;
int ans=0;
for (int i=1;i<=_m;i++) ans^=f[i];
return ans;
}

// BEGIN CUT HERE
public:
void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
private:
template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
void test_case_0() { int Arg0 = 0; int Arg1 = 3; int Arg2 = 3; verify_case(0, Arg2, count(Arg0, Arg1)); }
void test_case_1() { int Arg0 = 1; int Arg1 = 3; int Arg2 = 9; verify_case(1, Arg2, count(Arg0, Arg1)); }
void test_case_2() { int Arg0 = 3; int Arg1 = 3; int Arg2 = 73; verify_case(2, Arg2, count(Arg0, Arg1)); }
void test_case_3() { int Arg0 = 4; int Arg1 = 1; int Arg2 = 4; verify_case(3, Arg2, count(Arg0, Arg1)); }
void test_case_4() { int Arg0 = 70; int Arg1 = 39; int Arg2 = 319676671; verify_case(4, Arg2, count(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main(){
DerangementsStrikeBack ___test;
___test.run_test(-1);
getch() ;
return 0;
}
// END CUT HERE