【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 <iostream>
#include<bits/stdc++.h>
#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<j保证交换一次
//i做正常的+1，j左反转类型的+1,始终保持i和j是反转的
k = len/2;
while( j >= 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
,x2
;
int f
,a
;

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<len2;i++) x1[i]=Complex(f[i+l],0);
for(int i=len2;i<len;i++)   x1[i]=Complex(0,0);
for(int i=0;i<len1;i++) x2[i]=Complex(a[i],0);
for(int i=len1;i<len;i++)   x2[i]=Complex(0,0);
fft(x1,len,1);
fft(x2,len,1);
for(int i=0;i<len;i++)  x1[i]=x1[i]*x2[i];
fft(x1,len,-1);
for(int i=mid+1;i<=r;i++)   f[i]=(f[i]+(int)(x1[i-l].r+0.5))%MOD;
cdq(mid+1,r);
}

int main()
{
while(~scanf("%d",&n)&&n)
{
for(int i=1;i<=n;i++)   {scanf("%d",&a[i]);a[i]%=MOD;}
memset(f,0,sizeof(f));
cdq(1,n);
printf("%d\n",f
);
}
}