hdu5730 Shell Necklace

重温了这道cdq+FFT
讲白了就是不断对 dp[l~mid] 和 sh[1~r] 进行fft 得到 dp[mid+1~r]

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5+5;
const int MOD = 313;

int N;
int sh[MAXN], dp[MAXN];
int ans;
/****************FFT*************/
int A[MAXN<<2], B[MAXN<<2], C[MAXN<<2];
struct FFTSOLVE {
    int pos[MAXN<<2];
    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[MAXN<<2] , B[MAXN<<2] ;
    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 );
        }
    }
}boy;
void cdq(int l,int r) {
    if(l == r) {
        dp[l] = (dp[l] + sh[l]) % MOD;
        return;
    }
    int mid = (l+r)>>1;
    cdq(l,mid); 

    int l1 = 0, l2 = 0;
    for(int i = l; i <= mid; ++i) A[l1++] = dp[i];
    for(int i = 1; i <= N; ++i) {
        if(i+l > r) break;
        B[l2++] = sh[i];
    }
    int len = 1;
    while(len < l1*2 || len < l2*2) len <<= 1;
    for(int i = l1; i < len; ++i) A[i]=0;
    for(int i = l2; i < len; ++i) B[i]=0;
    boy.mul(A,B,C,len);
    for(int i = mid+1; i <= r; ++i) {
        dp[i] = (dp[i]+C[i-l-1]) %MOD;
    }
    cdq(mid+1,r);
}
int main(){
    while(~scanf("%d",&N)) {
        memset(dp,0,sizeof(dp));
        if(N == 0) break;
        for(int i = 1; i <= N; ++i) {
            scanf("%d",&sh[i]); sh[i] %= MOD;
        }

        cdq(1,N);
        printf("%d\n",dp[N]);
    }
    return 0;
}