Shell Necklace

有一个很显然的递推式:$dp[i]=\sum_{j=0}^{i-1} dp[j]a_{i-j}$,当然朴素的算法是$O(n^2)$,肯定过不了。观察这个式子,发现符合fft的形式(了解FFT),所以使用cdq分治+FFT(套路),复杂度为$O(nlog^{2}n)$。

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#include <bits/stdc++.h>
using namespace std;
const double PI=acos(-1.0);
const int MO=313;
const int N=100010;
struct Complex {
double x,y;
Complex(double _x = 0.0,double _y = 0.0)
{
x = _x;
y = _y;
}
Complex operator -(const Complex &b)const
{
return Complex(x-b.x,y-b.y);
}
Complex operator +(const Complex &b)const
{
return Complex(x+b.x,y+b.y);
}
Complex operator *(const Complex &b)const
{
return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
}
} x[280000],y[280000];
int a[N],dp[N];

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]);
k = len/2;
while(j >= k) {
j -= k;
k /= 2;
}
if(j < k)j += k;
}
}
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].x /= len;
}
}

void cdq(int l,int r)
{
if (l==r) {
dp[l]+=a[l];
dp[l]%=MO;
return;
}
int mid=(l+r)>>1;
cdq(l,mid);
int len1=mid-l+1;
int len2=r-l+1;
int len=1;
while (len<=len2) len<<=1;
for (int i=0;i<len1;i++) {
x[i]=Complex(dp[l+i],0);
}
for (int i=len1;i<len;i++) {
x[i]=Complex(0,0);
}
for (int i=0;i<len2;i++) {
y[i]=Complex(a[i],0);
}
for (int i=len2;i<len;i++) {
y[i]=Complex(0,0);
}
fft(x,len,1);
fft(y,len,1);
for (int i=0;i<len;i++) {
x[i]=x[i]*y[i];
}
fft(x,len,-1);
for (int i=mid+1;i<=r;i++) {
dp[i]+=round(x[i-l].x);
dp[i]%=MO;
}
cdq(mid+1,r);
}

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