Luogu4199 万径人踪灭

暑假的时候状态不好，现在发现这东西贼水

顺便复兴一下 (FFT) 板子

Description

给定一个 (a,b) 字符串，求非子串回文子序列个数

(|s| le 10^5)

Solution

按照题面概括，答案应该可以写成：回文子序列数 (-) 回文子串个数

后面的随便做一下就行了，比如 (hash+) 二分或 (Manacher)

然后考虑前面的部分：

令 (f_{i}=sumlimits_{j<i} [s_j=s_{i imes 2-j}])

那么以 (i) 为对称半径的答案就是 (2^{f_i}-1)

设 (a=1,b=0) 那么 (1 imes 1=1,0 imes 0=0 imes 1=1 imes 0=0) 卷积上去即可

再令 (a=0,b=1) 做一遍一样的

Code

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define reg register
namespace yspm{
	inline int read()
	{
		int res=0,f=1; char k;
		while(!isdigit(k=getchar())) if(k=='-') f=-1;
		while(isdigit(k)) res=res*10+k-'0',k=getchar();
		return res*f;
	}
	const int N=2e6+10;
	int ans[N],r[N],n,m;
	const double pi=acos(-1.0);
	struct node{
		double x,y;
		node(){}
		node(double xx,double yy){x=xx,y=yy; return ;}
		node operator+(node a){return node(a.x+x,y+a.y);}
		node operator-(node b){return node(x-b.x,y-b.y);}
		node operator*(node a){return node(x*a.x-y*a.y,x*a.y+y*a.x);} 
	}A[N];
	const int mod=1e9+7;
	inline int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
	inline int del(int x,int y){return x-y<0?x-y+mod:x-y;}
	inline int mul(int x,int y){return x*y-x*y/mod*mod;}
	inline int ksm(int x,int y)
	{
		int res=1;
		for(;y;y>>=1,x=mul(x,x)) if(y&1) res=mul(res,x); 
		return res;
	}
	inline void fft(node *f,int n,int opt)
	{	
		for(reg int i=0;i<n;++i) if(i<r[i]) swap(f[i],f[r[i]]);
		for(reg int p=2;p<=n;p<<=1)
		{
			int len=p>>1; node tmp(cos(pi/len),opt*sin(pi/len));
			for(reg int k=0;k<n;k+=p)
			{
				node buf(1,0);
				for(reg int l=k;l<k+len;++l,buf=buf*tmp)
				{
					node tt=buf*f[l+len];
					f[len+l]=f[l]-tt;
					f[l]=f[l]+tt;  
				}
			} 
		}
		if(opt==-1) for(reg int i=0;i<n;++i) f[i].x/=n;
		return ;
	}
	#define ull unsigned long long
	struct str{
		char s[N];
		int n,cnt,rmax,mid,len[N];
		inline int calc(char *t)
		{
			s[0]=s[1]='#'; int res=0;
			for(reg int i=1;i<=n;++i) s[i<<1]=t[i],s[i<<1|1]='#';
			n=(n+1)<<1; rmax=mid=r[1]=1;
			for(reg int i=2;i<n;++i)
			{
				if(i<rmax) r[i]=min(r[mid]+mid-i,r[(mid<<1)-i]); else r[i]=1;
				while(s[i-r[i]]==s[i+r[i]]) ++r[i];
				if(i+r[i]>rmax) mid=i,rmax=i+r[i];
				res=add(res,r[i]>>1);
			}
			return res; 
		} 
	}T;
	char s[N];
	signed main()
	{
		scanf("%s",s+1); 
		int len=strlen(s+1); T.n=len;
		int m=1,sum=0;
		while(m<=((len+1)<<1)) m<<=1;
		for(reg int i=0;i<m;++i) r[i]=r[i>>1]>>1|((i&1)?(m>>1):0);
		for(reg int i=1;i<=len;++i) if(s[i]=='a') A[i].x=1;
		fft(A,m,1); for(reg int i=0;i<m;++i) A[i]=A[i]*A[i]; fft(A,m,-1); 
		for(reg int i=0;i<m;++i) ans[i]+=(int)(A[i].x+0.5);
		memset(A,0,sizeof(A));
		for(reg int i=1;i<=len;++i) if(s[i]=='b') A[i].x=1;
		fft(A,m,1); for(reg int i=0;i<m;++i) A[i]=A[i]*A[i]; fft(A,m,-1);
		for(reg int i=0;i<m;++i) ans[i]+=(int)(A[i].x+0.5);
		for(reg int i=0;i<m;++i) sum=add(sum,del(ksm(2,(ans[i]+1)>>1),1));
		printf("%lld
",del(sum,T.calc(s)));
		return 0;
	}
}
signed main(){return yspm::main();}