LOJ #6051. 「雅礼集训 2017 Day11」PATH

完了感觉最近留了好多坑的说,这题也是模模糊糊地会一点

首先我们发现题目要求的是单调不上升的序列个数,那么一个套路就是用值减去下标

然后考虑连续位置的限制,这个我们做一个置换然后尽量向后取

这样拿值和位置卷积就变成了合法方案数的分子\(\times\)总方案数的分母?

感觉策不太懂啊,QQ上加了一个dalao问下细节,具体的到时候来填吧

先把CODE放了

#include<cstdio>
#include<cctype>
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
const int N=500005,mod=1004535809;
int n,m,len,lim,a[N],b[N],fact[N],F[N<<3],G[N<<3],ans;
class FileInputOutput
{
	private:
		static const int S=1<<21;
		#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
		char Fin[S],*A,*B;
	public:
		Tp inline void read(T& x)
		{
			x=0; char ch; while (!isdigit(ch=tc()));
			while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
		}
		#undef tc
}File;
inline void maxer(int& x,CI y)
{
	if (y>x) x=y;
}
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
	for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
class Poly_solver
{
	private:
		int rev[N<<3],p;
		inline void swap(int& x,int& y)
		{
			int t=x; x=y; y=t;
		}
		inline int sum(CI a,CI b)
		{
			int t=a+b; return t>=mod?t-mod:t;
		}
		inline int sub(CI a,CI b)
		{
			int t=a-b; return t<0?t+mod:t;
		}
	public:
		inline void init(CI n)
		{
			for (lim=1,p=0;lim<=n;lim<<=1,++p);
			for (RI i=0;i<lim;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<p-1);
		}
		inline void NTT(int *f,CI opt)
		{
			RI i; for (i=0;i<lim;++i) if (i<rev[i]) swap(f[i],f[rev[i]]);
			for (i=1;i<lim;i<<=1)
			{
				int m=i<<1,D=quick_pow(3,~opt?(mod-1)/m:mod-1-(mod-1)/m);
				for (RI j=0;j<lim;j+=m)
				{
					int W=1; for (RI k=0;k<i;++k,W=1LL*W*D%mod)
					{
						int x=f[j+k],y=1LL*f[i+j+k]*W%mod;
						f[j+k]=sum(x,y); f[i+j+k]=sub(x,y);
					}
				}
			}
			if (!~opt)
			{
				int Inv=quick_pow(lim); for (i=0;i<lim;++i) f[i]=1LL*f[i]*Inv%mod;
			}
		}
}P;
int main()
{
	//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
	RI i; for (File.read(n),i=1;i<=n;++i)
	File.read(a[i]),b[a[i]]=i,++F[a[i]-i+n];
	for (i=m=a[1];i;--i) maxer(b[i],b[i+1]),++G[b[i]-i+m];
	for (P.init((len=n+m)<<1),P.NTT(F,1),P.NTT(G,1),i=0;i<lim;++i)
	F[i]=1LL*F[i]*G[i]%mod; for (P.NTT(F,-1),i=ans=1;i<=len;++i)
	ans=1LL*ans*quick_pow(i,F[i-1+n+m])%mod; ans=quick_pow(ans);
	for (fact[0]=i=1;i<=m;++i) fact[i]=1LL*fact[i-1]*i%mod;
	for (i=1;i<=n;++i) ans=1LL*ans*fact[a[i]]%mod;
	return printf("%d",ans),0;
}
posted @ 2019-04-11 18:04  hl666  阅读(...)  评论(...编辑  收藏