BZOJ4517: [Sdoi2016] permutation count

Description How many sequence a of length n, satisfies the following conditions: 1 ~ n the number of n is present in the sequence once, if the number of I a[i] is the value of I, then I is said to be stable. Sequences that have exactly m number are stable enough to satisfy the condition of the sequence may be many, sequence number pairs 10^9+7 modulo. The first line of input is a number T, which indicates that there is a T group of data. Next T line, two integers per line n, M. t=500000,n≤1000000,m≤1000000 Output

Outputs T line, one number per line, indicating the number of sequences to be calculated

Sample Input5
1 0
1 1
5 2
100 50
10000Sample Output0
1
20
578028887
60695423first enumerate what numbers satisfy the ai=i, then it becomes the classic error-line problem. The recursive f[n]= (F[n-1]+f[n-2]) * (n-1) of the wrong row problem. We consider the number of the smallest person in the n person I, set him up to position J. 1. If the J person stands at position I, the problem turns into a n-2 problem. 2. If the J person does not stand in the I position, then except the I number of people all stand wrong, the problem turns into a n-1 problem of the wrong row. preprocessing factorial is the inverse element.

#include <cstdio> #include <cctype> #include <queue> #include <cstring> #include <algorithm > #define REP (i,s,t) for (int. i=s;i<=t;i++) #define DWN (I,S,T) for (int. i=s;i>=t;i--) #define REN for (int i=first[x ];i;i=next[i]) using namespace Std;const int Buffersize=1<<16;char Buffer[buffersize],*head,*tail;inline Char Getchar () {if (head==tail) {int l=fread (Buffer,1,buffersize,stdin); tail= (Head=buffer) +l;} return *head++;}    inline int read () {int X=0,f=1;char c=getchar (); for (;!    IsDigit (c); C=getchar ()) if (c== '-') f=-1;    for (; IsDigit (c); C=getchar ()) x=x*10+c-' 0 '; return x*f;} typedef long LONG ll;const int maxn=1000010;const int mod=1000000007;int xp[maxn],inv[maxn],f[maxn];int C (int n,int m) {RE Turn (ll) Xp[n]*inv[m]%mod*inv[n-m]%mod;} void init (int n) {xp[0]=inv[0]=inv[1]=1;rep (i,1,n) xp[i]= (LL) xp[i-1]*i%mod;rep (i,2,n) inv[i]= (LL) inv[mod%i]* ( mod-mod/i)%mod;rep (i,1,n) inv[i]= (LL) inv[i-1]*inv[i]%mod;f[2]=f[0]=1;rep (i,3,n) f[i]= (LL) (F[i-1]+f[i-2]) * (i-1)% MoD;} InchT A[maxn],b[maxn];int Main () {int n=read (), M=0;rep (i,1,n) M=max (M,a[i]=read ()), B[i]=read (), Init (m), Rep (i,1,n) {if (b[ I]>a[i]) puts ("0"), Else printf ("%d\n", (LL) C (A[i],b[i]) *f[a[i]-b[i]]%mod);} return 0;}

　　

BZOJ4517: [Sdoi2016] permutation count