#8 //HDU 5730 Shell Necklace（CDQ分治+FFT）

Description
给出长度分别为1~n的珠子，长度为i的珠子有a[i]种，每种珠子有无限个，问用这些珠子串成长度为n的链有多少种方案

题解：

• dp[i]表示组合成包含i个贝壳的项链的总方案数
• 转移：dp[i]=Σdp[i-j]*a[j]（1<=j<=i）

• #include <bits/stdc++.h>
  using namespace std;
  #define dob complex<double>
  #define rint register int
  #define mo 313
  #define IL inline
  const double pi=acos(-1.0);
  const int N=2e5;
  dob a[N],b[N],bb[N];
  int n,m,r[N],l,dp[N],sum[N];
  IL void fft(dob *a,int o)
  {
    for (rint i=;i<n;i++)
      if (i>r[i]) swap(a[i],a[r[i]]);
    for (rint i=;i<n;i*=)
    {
      dob wn(cos(pi/i),sin(pi*o/i)),x,y;
      for (rint j=;j<n;j+=(i*))
      {
        dob w(,);
        for (rint k=;k<i;k++,w*=wn)
        {
          x=a[j+k],y=w*a[i+j+k];
          a[j+k]=x+y,a[i+j+k]=x-y;
        }
      }
    }
  }
  IL void query()
  {
    l=;
    for (n=;n<=m;n<<=) l++;
    for (rint i=;i<n;i++) r[i]=(r[i/]/)|((i&)<<(l-));
    fft(a,); fft(b,);
    for (rint i=;i<n;i++) a[i]*=b[i];
    fft(a,-);
    for (rint i=;i<=m;i++)
      sum[i]=a[i].real()/n+0.5,sum[i]%=mo;
  }
  #define mid (l+r)/2
  void cdq(int l,int r)
  {
    if (l==r) return;
    cdq(l,mid);
    for (rint i=l;i<=mid;i++) a[i-l]=dp[i];
    m=r-l;
    rint x;
    for (x=;x<=m;x<<=);
    for (rint i=mid+;i<=l+x;i++) a[i-l]=;
    b[]=;
    for (rint i=;i<=x;i++) b[i]=bb[i];
    query();
    for (rint i=mid-l+;i<=r-l;i++)
      {
        dp[i+l]+=sum[i];
        dp[i+l]%=mo;
      }
    cdq(mid+,r);
  }
  int main()
  {
    freopen("noi.in","r",stdin);
    freopen("noi.out","w",stdout);
    std::ios::sync_with_stdio(false);
    int k;
    while (cin>>k&&k)
    {
      for (rint i=;i<=k;i++) cin>>bb[i];
      memset(dp,,sizeof(dp));
      dp[]=;
      cdq(,k);
      cout<<dp[k]%mo<<endl;
    }
    return ;
  }

该改一个fft模板了，实在是慢https://www.luogu.org/record/show?rid=3767323