HDU - 5730 ：Shell Necklace（CDQ分治+FFT）

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. 

InputThere are multiple test cases(no more than 2020 cases and no more than 1 in extreme case), ended by 0. 

For each test cases, the first line contains an integer nn, meaning the number of shells in this shell necklace, where 1≤n≤1051≤n≤105. Following line is a sequence with nnnon-negative integer a1,a2,…,ana1,a2,…,an, and ai≤107ai≤107 meaning the number of schemes to decorate ii continuous shells together with a declaration of love. 
OutputFor each test case, print one line containing the total number of schemes module 313313(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

        
 

Hint

For the first test case in Sample Input, the Figure 1 provides all schemes about it. The total number of schemes is 1 + 3 + 3 + 7 = 14.

 

见：https://blog.csdn.net/Maxwei_wzj/article/details/79850756

题意：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 
思路：对于一串珠子，列出DP方程，dp[i]表示长度为i的项链，所有情况的贡献和 dp[i]=∑ dp[j]*A[i-j]，复杂度O（N^2）。
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 。

（正确性不难证明，因为分治到Mid+1之前，dp[1]-dp[Mid]的值都已经求出来了。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=100010;
const int Mod=313;
const double pi=acos(-1.0);
struct cp
{
    double r,i;
    cp(){}
    cp(double rr,double ii):r(rr),i(ii){}
    cp operator +(const cp&x)const{return cp(r+x.r,i+x.i);}
    cp operator -(const cp&x)const{return cp(r-x.r,i-x.i);}
    cp operator *(const cp&x)const{return cp(r*x.r-i*x.i,i*x.r+r*x.i);}
};
ll dp[maxn],A[maxn];
cp a[maxn<<2],b[maxn<<2],W,w,p; int R[maxn<<2],n;
inline void fft(cp*c,int t)
{
    int i,j,k;
    for(i=0;i<n;i++) R[i]<i?swap(c[R[i]],c[i]),0:0;
    for(i=1;i<n;i<<=1)
     for(j=0,W={cos(pi/i),sin(pi/i)*t};j<n;j+=i<<1)
      for(k=0,w={1,0};k<i;k++,w=w*W)
       p=c[j+k+i]*w,c[j+k+i]=c[j+k]-p,c[j+k]=c[j+k]+p;
}
void solve(int l,int r)
{
    if(l==r) return ;
    int mid=(l+r)>>1 ;
    solve(l,mid);
    for(n=1;n<((r-l+1));n<<=1);
    rep(i,1,n-1) R[i]=R[i>>1]>>1|(i&1?n>>1:0);

    rep(i,0,mid-l) a[i]=cp(dp[l+i],0.);
    rep(i,mid-l+1,n) a[i]=cp(0,0);

    rep(i,0,r-l-1) b[i]=cp(A[i+1],0.);
    rep(i,r-l,n) b[i]=cp(0,0);

    fft(a,1); fft(b,1);
    rep(i,0,n-1) a[i]=a[i]*b[i];

    fft(a,-1);
    rep(i,mid+1,r) dp[i]+=a[i-l-1].r/n+0.5,dp[i]%=Mod;

    solve(mid+1,r);
}
int main()
{
    int N;
    while(~scanf("%d",&N)&&N>0){
        rep(i,1,N) dp[i]=0; dp[0]=1;
        rep(i,1,N) scanf("%d",&A[i]),A[i]%=Mod;
        solve(0,N);
        printf("%d\n",dp[N]);
    }
    return 0;
}