HDU 5730 多校1 Shell Necklace （CDQ分治+FFT）

Shell Necklace

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 694    Accepted Submission(s): 310

Problem 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 than20 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 module313(Three hundred and thirteen implies the march 13th, a special and purposeful day).

Sample Input

31 3 742 2 2 2 0

Sample Output

1454
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.
 

Author

HIT

Source

2016 Multi-University Training Contest 1

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题意：给你一段长为 i 的项链有 a[i] 种装饰方法，问长度为n的项链有多少种装饰方式。

题解：令dp[i]表示用这些珠子串成长度为i的链的方案数，并令dp[0]=1，可以得到转移方程 
 
由上式暴力求dp[n]时间复杂度O(n^2)，显然不行，数量级在1e5，考虑到上式右边是一个卷积形式，所以用CDQ分治+FFT来降低复杂度。

假设CDQ(l,r)为求出dp[l],dp[l+1],…,dp[r]的值，那么如果已经通过CDQ(l,mid)求出了dp[l],dp[l+1],…,dp[mid]，下面考虑dp[l],dp[l+1],…,dp[mid]对dp[mid+1],dp[mid+2],…,dp[r]的贡献。

令g[i]表示dp[l],…,dp[mid]对dp[i]的贡献，那么有，令x[i]=dp[i+l] (i=0,…,mid-l)，y[i]=a[i+1] (i=0,…,r-l-1)，则有 
 
所以对x序列和y序列做一遍FFT即可得到z序列，进而得到g序列 
总时间复杂度O(n*logn*logn) 。

AC代码：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>typedef long long ll;using namespace std;const int N = 2e5+10;const double pi = acos(-1.0);const int mod=313; char s1[N],s2[N];int len,res[N];struct Complex{    double r,i;    Complex(double r=0,double i=0):r(r),i(i) {};    Complex operator+(const Complex &rhs)    {        return Complex(r + rhs.r,i + rhs.i);    }    Complex operator-(const Complex &rhs)    {        return Complex(r - rhs.r,i - rhs.i);    }    Complex operator*(const Complex &rhs)    {        return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);    }} va[N],vb[N];//雷德算法--倒位序  void rader(Complex F[],int len) //len = 2^M,reverse F[i] with  F[j] j为i二进制反转{    int j = len >> 1;    for(int i = 1;i < len - 1;++i)    {        if(i < j) swap(F[i],F[j]);  // reverse        int k = len>>1;        while(j>=k)        {            j -= k;            k >>= 1;        }        if(j < k) j += k;    }}//FFT实现void FFT(Complex F[],int len,int t){    rader(F,len);    for(int h=2;h<=len;h<<=1) //分治后计算长度为h的DFT     {        Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h)); //单位复根e^(2*PI/m)用欧拉公式展开        for(int j=0;j<len;j+=h)        {            Complex E(1,0); //旋转因子            for(int k=j;k<j+h/2;++k)            {                Complex u = F[k];                Complex v = E*F[k+h/2];                F[k] = u+v; //蝴蝶合并操作                F[k+h/2] = u-v;                E=E*wn; //更新旋转因子            }        }    }    if(t==-1)   //IDFT        for(int i=0;i<len;++i)            F[i].r/=len;}//数组开小会RE ll a[4*N]; int b[4*N];  int dp[N]; Complex x[4*N],y[4*N];int n;//求卷积 void Conv(Complex a[],Complex b[],int len) {    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);}void gao(int l,int r)  {  if(l==r){dp[l]+=a[l];dp[l]%=mod;return ;}int mid=(l+r)>>1;gao(l,mid);int len=1;while(len<=r-l+1) len<<=1;for(int i=0;i<len;i++){if(i+l<=mid) x[i]=Complex(dp[i+l],0);        else x[i]=Complex(0,0);                if(l+i+1<=r) y[i]=Complex(a[i+1],0);        else y[i]=Complex(0,0);}    Conv(x,y,len);    for(int i=0;i<len;i++)    b[i]=(ll)(x[i].r+0.5)%mod;        for(int i=mid+1;i<=r;i++)    {    dp[i]+=b[i-l-1];    dp[i]%=mod;}gao(mid+1,r);}  void solve()  {      while(scanf("%d",&n),n)    {    memset(dp,0,sizeof(dp));    for(int i=1;i<=n;i++)    {    scanf("%d",&a[i]);    a[i]%=mod;}gao(1,n);printf("%d\n",dp[n]);}}  int main(){     solve();    return 0;}