hdu 5730 Shell Necklace(2016多校第一场)FFT+分治

dp[i]=sigma(dp[j]*a[i-j])

具有多项式形式可以考虑fft，但是直接会T，考虑cdq分治，每次分治时跑fft即可

#include 
#include
#include
#include
#include
using namespace std ;

typedef long long LL ;

#define clr( a , x ) memset ( a , x , sizeof a )

const int N = 524300 , P = 313 , M = 1000 ;
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 ;
        }
    }
}
const int mod=313;
const int maxn=200005;
int dp[maxn];
int a[maxn];
int b[maxn],c[maxn],d[maxn];
void cdq(int l,int r,int n){
    if(l==r){
        dp[l]+=a[l];
        dp[l]%=mod;
        return;
    }
    int mid=(l+r)/2;
    cdq(l,mid,n);
    int w=1;
    while(w<=r-l+1) w=w*2;
    for(int i=0;i