HDU5730 Shell Necklace（DP + CDQ分治 + FFT）

题目

Source

http://acm.hdu.edu.cn/showproblem.php?pid=5730

Description

Perhaps the sea‘s definition of a shell is the pearl. However, in my view, a shell necklace with n beautiful shells contains the most sincere feeling for my best lover Arrietty, but even that is not enough.

Suppose the shell necklace is a sequence of shells (not a chain end to end). Considering i continuous shells in the shell necklace, I know that there exist different schemes to decorate the i shells together with one declaration of love.

I want to decorate all the shells with some declarations of love and decorate each shell just one time. As a problem, I want to know the total number of schemes.

Input

There are multiple test cases(no more than 20 cases and no more than 1 in extreme case), ended by 0.

For each test cases, the first line contains an integer n, meaning the number of shells in this shell necklace, where 1≤n≤10⁵. Following line is a sequence with n non-negative integer a1,a2,…,an, and ai≤10⁷ meaning the number of schemes to decorate i continuous shells together with a declaration of love.

Output

For each test case, print one line containing the total number of schemes module 313(Three hundred and thirteen implies the march 13th, a special and purposeful day).

Sample Input

3
1 3 7
4
2 2 2 2
0

Sample Output

14
54

分析

题目大概说已知连续i（1<=i<=n）个贝壳组合成一段项链的方案数a[i]，求组合成包含n个贝壳的项链的总方案数。

• dp[i]表示组合成包含i个贝壳的项链的总方案数
• 转移：dp[i]=Σdp[i-j]*a[j]（1<=j<=i）

直接枚举转移的话时间复杂度O(n²)，是不行的。

其实这个转移方程是个比较特殊的卷积形式，可以用FFT去求，但是从1到n依次求的话时间复杂度是O(n²logn)。

而利用CQD分治，每次治的过程累加左半区间内各个已经求得dp值的状态对右半区间各个状态的贡献，这个贡献就是那个方程用FFT求即可，这样时间复杂度由主定理可知是O(nlog²n)。

这是个很经典的题目吧，虽然现在才做。。

代码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 333333
const double PI=acos(-1.0);
 
struct Complex{
	double real,imag;
	Complex(double _real,double _imag):real(_real),imag(_imag){}
	Complex(){}
	Complex operator+(const Complex &cp) const{
		return Complex(real+cp.real,imag+cp.imag);
	}
	Complex operator-(const Complex &cp) const{
		return Complex(real-cp.real,imag-cp.imag);
	}
	Complex operator*(const Complex &cp) const{
		return Complex(real*cp.real-imag*cp.imag,real*cp.imag+cp.real*imag);
	}
	void setValue(double _real=0,double _imag=0){
		real=_real; imag=_imag;
	}
};
 
int len;
Complex wn[MAXN],wn_anti[MAXN];
 
void FFT(Complex y[],int op){
	for(int i=1,j=len>>1,k; i<len-1; ++i){
		if(i<j) swap(y[i],y[j]);
		k=len>>1;
		while(j>=k){
			j-=k;
			k>>=1;
		}
		if(j<k) j+=k;
	}
	for(int h=2; h<=len; h<<=1){
		Complex Wn=(op==1?wn[h]:wn_anti[h]);
		for(int i=0; i<len; i+=h){
			Complex W(1,0);
			for(int j=i; j<i+(h>>1); ++j){
				Complex u=y[j],t=W*y[j+(h>>1)];
				y[j]=u+t;
				y[j+(h>>1)]=u-t;
				W=W*Wn;
			}
		}
	}
	if(op==-1){
		for(int i=0; i<len; ++i) y[i].real/=len;
	}
}
void Convolution(Complex A[],Complex B[],int n){
	for(len=1; len<(n<<1); len<<=1);
	for(int i=n; i<len; ++i){
		A[i].setValue();
		B[i].setValue();
	}
	
	FFT(A,1); FFT(B,1);
	for(int i=0; i<len; ++i){
		A[i]=A[i]*B[i];
	}
	FFT(A,-1);
}

int a[111111],d[111111];
Complex A[MAXN],B[MAXN];

void cdq(int l,int r){
	if(l==r){
		d[l]+=a[l];
		d[l]%=313;
		return;
	}
	int mid=l+r>>1;
	cdq(l,mid);

	for(int i=l; i<=mid; ++i) A[i-l].setValue(d[i]);
	for(int i=0; i<=r-l; ++i) B[i].setValue(a[i]);
	for(int i=mid-l+1; i<=r-l; ++i) A[i].setValue();
	Convolution(A,B,r-l+1);
	for(int i=mid+1; i<=r; ++i){
		d[i]+=((long long)(A[i-l].real+0.5))%313;
		d[i]%=313;
	}
	
	cdq(mid+1,r);
}

int main(){
	for(int i=0; i<MAXN; ++i){
		wn[i].setValue(cos(2.0*PI/i),sin(2.0*PI/i));
		wn_anti[i].setValue(wn[i].real,-wn[i].imag);
	}
	int n;
	while(~scanf("%d",&n) && n){
		for(int i=1; i<=n; ++i){
			scanf("%d",a+i);
			a[i]%=313;
		}
		memset(d,0,sizeof(d));
		cdq(1,n);
		printf("%d
",d[n]);
	}
	return 0;
}