HDU 5730 Shell Necklace(dp+cdq分治+FFT优化)

一串项链是n个珠子组成，如果i个珠子连续，可以被认为是模式i，贡献是ai

对于一串珠子，如果用了模式b1,b2,...bk，贡献就是∏mi=1abi

求n长度的项链，所有情况的贡献和

可以列出dp方程，f[i]表示长度为i的项链，所有情况的贡献和

f[i]=∑ij=1f[i−j]×aj

O(n)的状态，O(n)的转移，总复杂度O(n2)，GG

观察这个式子很像卷积的形式，于是可以用cdq分治+FFT来优化一波

先求出前面的f的值，然后去算后面的f的值，分治logn复杂度

然后每次分治的时候，需要nlogn的FFT计算

总复杂度O(nlog2n)

代码：

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           100005
#define   MAXN          1000005
#define   maxnode       105
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-6;
const LL     mod   = 313;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const double PI = acos(-1.0);
struct Complex  {
double real, image;
Complex(double _real, double _image)  {
real = _real;
image = _image;
}
Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)  {
return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)  {
return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)  {
return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len)  {
int ret = 0;
for(int i = 0; (1 << i) < len; i++) {
ret <<= 1;
if(id & (1 << i)) ret |= 1;
}
return ret;
}

Complex A[280000];  /////////大小要修改，不要忘记

void FFT(Complex* a, int len, int DFT){//对a进行DFT或者逆DFT, 结果存在a当中
//Complex* A = new Complex[len]; 这么写会爆栈
for(int i = 0; i < len; i++)
A[rev(i, len)] = a[i];
for(int s = 1; (1 << s) <= len; s++)  {
int m = (1 << s);
Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
for(int k = 0; k < len; k += m) {
Complex w = Complex(1, 0);
for(int j = 0; j < (m >> 1); j++) {
Complex t = w*A[k + j + (m >> 1)];
Complex u = A[k + j];
A[k + j] = u + t;
A[k + j + (m >> 1)] = u - t;
w = w*wm;
}
}
}
if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
for(int i = 0; i < len; i++) a[i] = A[i];
return;
}

Complex a[280000], b[280000];//对应的乘积的长度不会超过100000, 也就是不超过(1 << 17) = 131072
int f[MAX];
int x[MAX];
void cdq(int l,int r){
if(l==r) return;
int mid=(l+r)/2;
cdq(l,mid);
int len=1;
while(len<=r-l+1) len<<=1;
for(int i=0;i<len;i++){
if(l+i<=mid) a[i]=Complex(f[l+i],0);
else a[i]=Complex(0,0);
if(l+i+1<=r) b[i]=Complex(x[i+1],0);
else b[i]=Complex(0,0);
}
FFT(a,len,1);
FFT(b,len,1);
for(int i=0;i<len;i++) a[i]=a[i]*b[i];
FFT(a,len,-1);
for(int i=mid+1;i<=r;i++){
f[i]+=floor(a[i-l-1].real+0.5);
f[i]%=mod;
}
cdq(mid+1,r);
}

int main(){
//freopen("in.txt","r",stdin);
int n;
while(cin>>n&&n){
for(int i=1;i<=n;i++) scanf("%d",&x[i]),x[i]%=mod;
mem(f,0);
f[0]=1;
cdq(0,n);
printf("%d\n",f
);
}
return 0;
}