HDU 5730 Shell Necklace (CDQ Division +fft)

"Topic link" http://acm.hdu.edu.cn/showproblem.php?pid=5730

"The main topic"

Give An array w, representing the weights of the fields of different lengths, such as W[3]=5, if the field length is 3, then its weight is 5, and now there is a field of length n, the sum of the values of the field weights obtained by different splits.

Exercises

　　Note Dp[i] is the answer to the length of I, dp[i]=sum_{j=0}^{i-1}dp[j]w[i-j], found to be a convolution of the formula, so the operation can be optimized by FFT, but because in the calculation process Dp[j] is an unknown value, the sequential computational complexity is O ( N2LOGN), considering that the addition operation can be allocated to the multiplication operation, it is possible to take CDQ division, using recursive statistics to the left DP value in each interval contribution to the right DP value, the calculation of each contribution value is optimized by FFT, optimizing the time complexity to O (NLOGNLOGN).

Code

#include <cstdio> #include <cmath> #include <algorithm>using namespace std;typedef long Long ll;const        int n=524300,p=313;int n,pos[n];namespace fft{struct comp{double r,i;        Comp (double _r=0,double _i=0): R (_r), I (_i) {} comp operator + (const comp&x) {return comp (r+x.r,i+x.i);}        Comp operator-(const comp&x) {return comp (r-x.r,i-x.i);}        Comp operator * (const comp&x) {return comp (r*x.r-i*x.i,i*x.r+r*x.i);}    Comp Conj () {return comp (r,-i);}    }a[n],b[n];    Const double Pi=acos (-1.0);        void FFT (comp a[],int n,int t) {for (int i=1;i<n;i++) if (pos[i]>i) swap (a[i],a[pos[i]]);            for (int d=0; (1<<d) <n;d++) {int m=1<<d,m2=m<<1;            Double o=pi*2/m2*t;            Comp _w (cos (o), sin (o));                for (int i=0;i<n;i+=m2) {Comp W (1,0);                    for (int j=0;j<m;j++) {comp& a=a[i+j+m],&b=a[i+j],t=w*a;  A=b-t;                  B=b+t;                W=w*_w;    }}}if (T==-1) for (int i=0;i<n;i++) a[i].r/=n;        } void Mul (int *a,int *b,int *c,int k) {int i,j;        for (i=0;i<k;i++) A[i]=comp (A[i],b[i]);        J=__builtin_ctz (k)-1; for (int i=0;i<k;i++) {pos[i]=pos[i>>1]>>1| (         (i&1) <<j);}        FFT (a,k,1);            for (int i=0;i<k;i++) {j= (k-i) & (K-1);        B[i]= (a[i]*a[i]-(A[j]*a[j]). CONJ ()) *comp (0,-0.25);        }fft (b,k,-1);    for (int i=0;i<k;i++) c[i]= (Long Long) (b[i].r+0.5)%P; }}int w[n],a[n],b[n],c[n],f[n];void CDQ (int l,int r) {if (l==r) {f[l]+=w[l];    F[l]%=p;return;}    int mid= (L+R) >>1; CDQ (L,mid);    int n=1;    while (n<r-l) n<<=1;    for (int i=0;i<=mid-l;i++) a[i]=f[i+l];    for (int i=mid-l+1;i<n;i++) a[i]=0;    for (int i=0;i<r-l;i++) b[i]=w[i+1];    for (int i=r-l;i<n;i++) b[i]=0;    Fft::mul (A,b,c,n); for (int i=mid+1;i<=r;i++) {f[i]+=c[i-l-1];    F[i]%=p; }CDQ (mid+1,r);}        int main () {while (~SCANF ("%d", &n) &&n) {f[0]=1;            for (int i=1;i<=n;i++) {scanf ("%d", &w[i]);            W[i]%=p;        f[i]=0;        }CDQ (1,n);    printf ("%d\n", F[n]); }return 0;}

　　

HDU 5730 Shell Necklace (CDQ Division +fft)