hdu 5730 - CDQ分治 + fft
阿新 • • 發佈:2018-10-31
題目連結:點選這裡
解題思路:
dp[]表示長度為i的項鍊的方案值.
dp[i] = ∑dp[j]*a[i-j] (j<i)
兩邊都有dp[],所以選擇分治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; }