【多校训练】hdu 5730 cdq+fft

Shell Necklace

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1089    Accepted Submission(s): 467

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 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  nnon-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

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.

这题很容易就推出公式    

f[1]=0; 

f[i]=∑(f[i - j] * a[j]), j∈[1, i];

但是如果按照公式暴力做的话会超时。看到这种形式的式子应该想到fft（...其实我也没想到，之前没怎么看过fft，趁机会学习了一下。

但是做n次fft是会超时的，所以还需要用到cdq二分的方法来优化，cdq二分之前没见过，学习了一下，感觉挺巧妙的。。。

总之通过cdq+fft，这题也就出来了。其实没有什么思维上的难点，主要的对两个算法的了解（套模板，但是如果不熟悉的话还是很难做出的。

#include 
#include
#define ll long long
using namespace std;
const int MOD=313;
const int N=410000;
int n;

const double PI = acos(-1.0);
//复数结构体
struct Complex
{
    double r,i;
    Complex(double _r = 0.0,double _i = 0.0)
    {
        r = _r; i = _i;
    }
    Complex operator +(const Complex &b)
    {
        return Complex(r+b.r,i+b.i);
    }
    Complex operator -(const Complex &b)
    {
        return Complex(r-b.r,i-b.i);
    }
    Complex operator *(const Complex &b)
    {
        return Complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂
 */
void change(Complex y[],int len)
{
    int i,j,k;
    for(i = 1, j = len/2;i < len-1; i++)
    {
        if(i < j)swap(y[i],y[j]);
        //交换互为小标反转的元素，i= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
/*
 * 做FFT
 * len必须为2^k形式，
 * on==1时是DFT，on==-1时是IDFT
 */
void fft(Complex y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j = 0;j < len;j+=h)
        {
            Complex w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                Complex u = y[k];
                Complex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].r /= len;
}

Complex x1[N],x2[N];
int f[N],a[N];

void cdq(int l, int r)
{
    if(l==r)
    {
        f[l]=(f[l]+a[l])%MOD;
        return ;
    }
    int mid=l+r >> 1;
    cdq(l,mid);
    int len1 = r - l + 1;
    int len2 = mid - l + 1;
    int len=1;
    while(len<(len1+len2))  len<<=1;
    for(int i=0;i