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  • Solution -「SDOI 2017」「洛谷 P3784」遗忘的集合

    (mathcal{Description})

      Link.

      给定 ({f_1,f_2,cdots,f_n}),素数 (p)。求字典序最小的 ({a_1,a_2,cdots,a_n}),满足对于所有 (iin[1,n])(a_iin{0,1}) 并且

    [sum_{{k_{1..n}}}[(forall j)left((a_j=0land k_j=0)lor(a_j ot=0land k_jge0 ight)]left[ sum_{j=1}^njk_j=i ight]equiv f_ipmod p ]

      (n<2^{18})

    (mathcal{Solution})

      对于某个 (a_i=1),其 OGF 为 (frac{1}{1-x^i}),所有 OGF 之积的 (1sim n) 次项系数在(mod p) 意义下就是 (f_{1..n})。记 (F(x)=sum_{i=0}^{+infty}f_ix^i),默认运算在(mod p) 意义下进行,那么:

    [F(x)=prod_{i=1}^nleft( frac{1}{1-x^i} ight)^{a_i} ]

      两边取 (-ln)

    [Rightarrow~~~~-ln F(x)=sum_{i=1}^na_iln(1-x^i) ]

      Tayler 展开右式 (ln)

    [Rightarrow~~~~-ln F(x)=sum_{i=1}^na_isum_{j=1}^{+infty}-frac{x^{ij}}j ]

      左右取负,枚举 (T=ij)

    [Rightarrow~~~~ln F(x)=sum_{T=1}^{+infty}x^Tsum_{i|T}a_ifrac{i}T ]

      对已知的 (F(x))(ln),就能取得 (Tin[1,n]) 内的所有 (frac{1}Tsum_{i|T}a_ii),从 (T=1) 起刷表求解,每次枚举倍数消除贡献,可以 (mathcal O(nln n)) 解出所有 (a_{1..n})(没错,由于 (ln F(x)) 模意义下的多项式系数唯一,({a_n}) 唯一确定)。

      综上,复杂度为多项式 (ln)(mathcal O(nlog n))。建议用毛爷爷的四次 FFT 科技做 MTT。

    (mathcal{Code})

    /* Clearink */
    
    #include <cmath>
    #include <cstdio>
    #include <cassert>
    
    #define rep( i, l, r ) for ( int i = l, rpbound##i = r; i <= rpbound##i; ++i )
    #define per( i, r, l ) for ( int i = r, rpbound##i = l; i >= rpbound##i; --i )
    
    typedef long long LL;
    
    inline int rint () {
    	int x = 0; char s = getchar ();
    	for ( ; s < '0' || '9' < s; s = getchar () );
    	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
    	return x;
    }
    
    inline void wint ( const int x ) {
    	if ( 9 < x ) wint ( x / 10 );
    	putchar ( x % 10 ^ '0' );
    }
    
    const int MAXLEN = 1 << 19;
    const double PI = acos ( -1 );
    int n, M, f[MAXLEN + 5], a[MAXLEN + 5];
    
    inline int mul ( const long long a, const int b ) { return a * b % M; }
    inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + M : a; }
    inline int add ( int a, const int b ) { return ( a += b ) < M ? a : a - M; }
    inline int mpow ( int a, int b ) {
    	int ret = 1;
    	for ( ; b; a = mul ( a, a ), b >>= 1 ) ret = mul ( ret, b & 1 ? a : 1 );
    	return ret;
    }
    
    namespace PolyOper {
    
    int rev[MAXLEN + 5], inv[MAXLEN + 5];
    
    inline void initInv () {
    	inv[1] = 1;
    	rep ( i, 2, MAXLEN ) inv[i] = mul ( M - M / i, inv[M % i] );
    }
    
    struct Complex {
    	double x, y;
    	Complex (): x ( 0 ), y ( 0 ) {}
    	Complex ( const double tx, const double ty ): x ( tx ), y ( ty ) {}
    	inline Complex operator + ( const Complex t ) const {
    		return Complex ( x + t.x, y + t.y );
    	}
    	inline Complex operator - ( const Complex t ) const {
    		return Complex ( x - t.x, y - t.y );
    	}
    	inline Complex operator * ( const Complex t ) const {
    		return Complex ( x * t.x - y * t.y, x * t.y + y * t.x );
    	}
    	inline Complex operator / ( const double t ) const {
    		return Complex ( x / t, y / t );
    	}
    };
    Complex omega[MAXLEN + 5], P[MAXLEN + 5], Q[MAXLEN + 5];
    Complex C[MAXLEN + 5], D[MAXLEN + 5], E[MAXLEN + 5], F[MAXLEN + 5];
    
    inline void FFT ( const int n, Complex* A, const int tp ) {
    	rep ( i, 0, n - 1 ) if ( i < rev[i] ) std::swap ( A[i], A[rev[i]] );
    	for ( int i = 2, stp = 1; i <= n; i <<= 1, stp <<= 1 ) {
    		for ( int j = 0; j < n; j += i ) {
    			rep ( k, 0, stp - 1 ) {
    				Complex w ( omega[n / stp * k].x, tp * omega[n / stp * k].y );
    				Complex ev ( A[j + k] ), ov ( w * A[j + k + stp] );
    				A[j + k] = ev + ov, A[j + k + stp] = ev - ov;
    			}
    		}
    	}
    	if ( !~tp ) rep ( i, 0, n - 1 ) A[i] = A[i] / n;
    }
    
    inline void initFFT ( const int lg ) {
    	int n = 1 << lg;
    	rep ( i, 0, n - 1 ) rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << lg >> 1 );
    	for ( int i = 1; i < n; i <<= 1 ) {
    		rep ( k, 0, i - 1 ) {
    			omega[n / i * k] = Complex ( cos ( PI * k / i ), sin ( PI * k / i ) );
    		}
    	}
    }
    
    inline void polyConv ( const int n, const int m, const int* A, const int* B, int* R ) {
    	rep ( i, 0, n - 1 ) P[i] = Complex ( A[i] & 0x7fff, A[i] >> 15 );
    	rep ( i, 0, m - 1 ) Q[i] = Complex ( B[i] & 0x7fff, B[i] >> 15 );
    	int lg = 0, len = 1;
    	for ( ; len < n + m - 1; len <<= 1, ++ lg );
    	rep ( i, n, len ) P[i] = Complex ();
    	rep ( i, m, len ) Q[i] = Complex ();
    	initFFT ( lg );
    	FFT ( len, P, 1 ), FFT ( len, Q, 1 );
    	rep ( i, 0, len - 1 ) {
    		Complex t ( P[( len - i ) % len].x, -P[( len - i ) % len].y );
    		C[i] = ( P[i] + t ) / 2, D[i] = Complex ( 0, 1 ) * ( t - P[i] ) / 2;
    	}
    	rep ( i, 0, len - 1 ) {
    		Complex t ( Q[( len - i ) % len].x, -Q[( len - i ) % len].y );
    		E[i] = ( Q[i] + t ) / 2, F[i] = Complex ( 0, 1 ) * ( t - Q[i] ) / 2;
    	}
    	rep ( i, 0, len - 1 ) {
    		Complex c ( C[i] ), d ( D[i] ), e ( E[i] ), f ( F[i] );
    		C[i] = c * e, D[i] = c * f + d * e, E[i] = d * f;
    		P[i] = C[i] + Complex ( 0, 1 ) * E[i];
    	}
    	FFT ( len, D, -1 ), FFT ( len, P, -1 );
    	rep ( i, 0, n + m - 2 ) {
    		int c = ( LL ( P[i].x + 0.5 ) % M + M ) % M,
    			d = ( LL ( D[i].x + 0.5 ) % M + M ) % M,
    			e = ( LL ( P[i].y + 0.5 ) % M + M ) % M;
    		R[i] = ( c + ( 1ll << 15 ) % M * d % M + ( 1ll << 30 ) % M * e % M ) % M;
    	}
    }
    
    inline void polyDer ( const int n, const int* A, int* R ) {
    	rep ( i, 1, n - 1 ) R[i - 1] = mul ( i, A[i] );
    	R[n - 1] = 0;
    }
    
    inline void polyInt ( const int n, const int* A, int* R ) {
    	per ( i, n - 1, 0 )	R[i + 1] = mul ( inv[i + 1], A[i] );
    	R[0] = 0;
    }
    
    inline void polyInv ( const int n, const int* A, int* R ) {
    	if ( n == 1 ) return void ( R[0] = mpow ( A[0], M - 2 ) );
    	static int tmp[MAXLEN + 5];
    	polyInv ( n + 1 >> 1, A, R ); polyConv ( n, n, A, R, tmp );
    	tmp[0] = sub ( 2, tmp[0] );
    	rep ( i, 1, n - 1 ) tmp[i] = sub ( 0, tmp[i] );
    	polyConv ( n, n, tmp, R, R );
    }
    
    inline void polyLn ( const int n, const int* A, int* R ) {
    	static int tmp[2][MAXLEN + 5];
    	polyDer ( n, A, tmp[0] ), polyInv ( n, A, tmp[1] );
    	polyConv ( n, n, tmp[0], tmp[1], tmp[0] );
    	polyInt ( n, tmp[0], R );
    }
    
    } // namesapce PolyOper.
    
    int main () {
    	n = rint (), M = rint ();
    	PolyOper::initInv ();
    	f[0] = 1;
    	rep ( i, 1, n ) f[i] = rint ();
    	PolyOper::polyLn ( n + 1, f, f );
    	rep ( i, 1, n ) f[i] = mul ( f[i], i );
    	int cnt = 0;
    	rep ( i, 1, n ) {
    		assert ( !f[i] || f[i] == i );
    		cnt += a[i] = !!f[i];
    		if ( a[i] ) rep ( j, 2, n / i ) f[i * j] = sub ( f[i * j], i );
    	}
    	wint ( cnt );
    	for ( int i = 1, flg = 0; i <= n; ++i ) {
    		if ( a[i] ) {
    			putchar ( flg ? ' ' : '
    ' ), flg = 1;
    			wint ( i );
    		}
    	}
    	putchar ( '
    ' );
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/rainybunny/p/14248981.html
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