hdu5730 Shell Necklace 【分治fft】

题目

简述：

有一段长度为n的贝壳，将其划分为若干段，给出划分为每种长度的方案数，问有多少种划分方案

题解

设\(f[i]\)表示长度为\(i\)时的方案数
不难得dp方程：

\[f[i] = \sum\limits_{j=0}^{i} a[j] * f[i - j] \]

考虑转移
直接转移是\(O(n^2)\)的
如何优化？
容易发现这个转移方程非常特别，是一个卷积的形式
考虑fft

分治fft##

分治fft解决的就是这样一个转移方程的快速计算的问题

\[f[i] = \sum\limits_{j=0}^{i} a[j] * f[i - j] \]

考虑cdq分治的模式：
我们先处理左半区间，然后用左半区间的\(f[i]\)来更新右半区间的答案
具体地，左半区间对右边一个位置\(r\)的贡献是：

\[\sum\limits_{i=l}^{mid} f[i] * a[r - i] \]

也是一个卷积的形式，为多项式乘积的第\(r\)项
如此我们便可以用\(f[i]\)和\(a[i]\)构造两个多项式，作fft，然后直接将相应位置的值累加到右半边相应位置的\(f[i]\)中去

我们便解决了这道题

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000,P = 313;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
const double pi = acos(-1);
struct E{
	double r,i;
	E(){}
	E(double a,double b):r(a),i(b){}
	E operator =(const int& b){
		r = b; i = 0;
		return *this;
	}
};
inline E operator +(const E& a,const E& b){
	return E(a.r + b.r,a.i + b.i);
}
inline E operator -(const E& a,const E& b){
	return E(a.r - b.r,a.i - b.i);
}
inline E operator *(const E& a,const E& b){
	return E(a.r * b.r - a.i * b.i,a.r * b.i + b.r * a.i);
}
inline E operator *=(E& a,const E& b){
	return (a = a * b);
}
inline E operator /(E& a,const double& b){
	return E(a.r / b,a.i / b);
}
inline E operator /=(E& a,const double& b){
	return (a = a / b);
}
int n,m,L,R[maxn];
void fft(E* a,int f){
	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){
		E wn(cos(pi / i),f * sin(pi / i));
		for (int j = 0; j < n; j += (i << 1)){
			E w(1,0);
			for (int k = 0; k < i; k++,w *= wn){
				E x = a[j + k],y = w * a[j + k + i];
				a[j + k] = x + y; a[j + k + i] = x - y;
			}
		}
	}
	if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
E A[maxn],B[maxn];
int N,a[maxn],f[maxn];
void solve(int l,int r){
	if (l == r){
		f[l] = (f[l] + a[l]) % P;
		return;
	}
	int mid = l + r >> 1;
	solve(l,mid);
	n = mid - l + 1;
	for (int i = 0; i < n; i++) A[i] = f[l + i];
	m = r - l + 1;
	for (int i = 0; i < m; i++) B[i] = a[i + 1];
	m = n + m; L = 0;
	for (n = 1; n <= m; n <<= 1) L++;
	for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	fft(A,1); fft(B,1);
	for (int i = 0; i < n; i++) A[i] *= B[i];
	fft(A,-1);
	for (int i = mid + 1; i <= r; i++){
		f[i] = (f[i] + (int)floor(A[i - l - 1].r + 0.5) % P) % P;
	}
	for (int i = 0; i < n; i++) A[i] = B[i] = 0;
	solve(mid + 1,r);
}
int main(){
	while ((~scanf("%d",&N)) && N){
		for (int i = 1; i <= N; i++){
			a[i] = read() % P;
			f[i] = 0;
		}
		solve(1,N);
		printf("%d\n",f[N]);
	}
	return 0;
}