hdu 5730 - CDQ分治 + fft

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题目链接:点击这里

解题思路:

dp[]表示长度为i的项链的方案值.

dp[i] = ∑dp[j]*a[i-j] (j<i)

两边都有dp[],所以选择分治FFT

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double pi = acos(-1);
const int mod = 313;
const int mx = 1<<18;
typedef complex<double> comp;
int n,rev[mx],dp[mx];
comp a[mx],c[mx],d[mx];
void get_rev(int len)
{
    for(int i=1;i<(1<<len);i++)
    rev[i] = (rev[i>>1]>>1)|((i&1)<<(len-1));
}
void fft(comp *p,int len,int v)
{
    for(int i=0;i<len;i++)
    if(i<rev[i]) swap(p[i],p[rev[i]]);
    for(int i=1;i<len;i<<=1) 
    {
        comp tep = exp(comp(0,v*pi/i));
        for(int j=0;j<len;j+=(i<<1))
        {
            comp base(1,0);
            for(int k=j;k<j+i;k++) 
            {
                comp x = p[k];  
                comp y = base*p[k+i]; 
                p[k] = x + y;
                p[k+i] =  x-y; 
                base *= tep;
            } 
        }
    }
    if(v==-1) for(int i=0;i<len;i++) p[i] /= len; 
}
void cdq(int l,int r,int bi)
{
    if(l==r){
        dp[l] = (dp[l]+int(a[l].real()))%mod;
        return ;
    }
    int mid = (l+r)>>1,len = (r-l+1);
    cdq(l,mid,bi-1);
    for(int i=0;i<len/2;i++) c[i] = comp(dp[i+l],0);
    for(int i=len/2;i<2*len;i++) c[i] = comp(0,0);
    for(int i=0;i<len;i++) d[i] = a[i+1];
    for(int i=len;i<2*len;i++) d[i] = comp(0,0);
    get_rev(bi);
    fft(c,2*len,1);
    fft(d,2*len,1);
    for(int i=0;i<2*len;i++) c[i] = c[i]*d[i];
    fft(c,2*len,-1);
    for(int i=mid+1;i<=r;i++) dp[i] = (dp[i]+int(c[i-l-1].real()+0.5))%mod;
    cdq(mid+1,r,bi-1);
}
int main()
{
    int u;
    while(scanf("%d",&n)&&n)
    {
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++){
            scanf("%d",&u);
            u %= mod;
            a[i] = comp(u,0);
        }
        int len = 1;
        while((1<<len)<n) len++;
        cdq(1,1<<len,len+1);
        printf("%d\n",dp[n]);
    }
    return 0;
}