为什么我这么弱

其实FFT也挺水的，一点数学基础加上细心即可。细节·技巧挺多。

递归

在TLE的边缘苦苦挣扎

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)

#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)

#define Max(a,b) ((a) > (b) ? (a) : (b))

#define Min(a,b) ((a) < (b) ? (a) : (b))

#define Fill(a,b) memset(a, b, sizeof(a))

#define Abs(a) ((a) < 0 ? -(a) : (a))

#define Swap(a,b) a^=b^=a^=b

#define ll long long

//#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")

#define D_e(x)  cout << #x << " = " << x << endl

#define Pause() system("pause")

#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;

#define D_e(x)  ;

#define Pause() ;

#define FileOpen() ;

#endif

struct ios{

    template<typename ATP>ios& operator >> (ATP &x){

        x = 0; int f = 1; char c;

        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;

        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();

        x*= f;

        return *this;

    }

}io;

using namespace std;

const int N = 4000007; // how much should I open QAQ ?

const double pi = acos(-1.0);

struct Complex{

	double x, y;

	Complex (double xx = 0, double yy = 0) {x = xx, y = yy;}

	Complex operator + (Complex b){ return Complex(x + b.x, y + b.y); }

	Complex operator - (Complex b){ return Complex(x - b.x, y - b.y); }

	Complex operator * (Complex b){ return Complex(x * b.x - y * b.y, x * b.y + y * b.x); }

}a[N], b[N];

inline void FFT(int limit, Complex *a, int opt){

	if(limit == 1) return;

	Complex a1[(limit >> 1) + 3], a2[(limit >> 1) + 3];

	for(register int i = 0; i <= limit; i += 2){

		a1[i >> 1] = a[i];

		a2[i >> 1] = a[i + 1];

	}

	FFT(limit >> 1, a1, opt);

	FFT(limit >> 1, a2, opt);

	Complex Wn = Complex( cos(2.0 * pi / limit), opt * sin(2.0 * pi / limit));

	Complex w = Complex( 1, 0);

	R(i,0,(limit >> 1) - 1){ // be careful, do not write 'R(i,0,(limit >> 1))'

		a[i] = a1[i] + w * a2[i];

		a[i + (limit >> 1)] = a1[i] - w * a2[i];

		w = w * Wn;

	}

}

int main(){

	int n, m;

	io >> n >> m;

	R(i,0,n) io >> a[i].x;

	R(i,0,m) io >> b[i].x;

	int limit;

	for(limit = 1; limit <= n + m; limit <<= 1);

	FFT(limit, a, 1);

	FFT(limit, b, 1); // coefficient changes to point value

	R(i,0,limit){

		a[i] = a[i] * b[i];

	}

	FFT(limit, a, -1); // point value changes to coefficient

	R(i,0,n + m){

		printf("%d ", (int)(a[i].x / limit + 0.5)); // ans should divide limit

	}

	return 0;

}

迭代

快得飞起\ *^* /

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)

#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)

#define Max(a,b) ((a) > (b) ? (a) : (b))

#define Min(a,b) ((a) < (b) ? (a) : (b))

#define Fill(a,b) memset(a, b, sizeof(a))

#define Abs(a) ((a) < 0 ? -(a) : (a))

#define Swap(a,b) a^=b^=a^=b

#define ll long long

//#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")

#define D_e(x)  cout << #x << " = " << x << endl

#define Pause() system("pause")

#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;

#define D_e(x)  ;

#define Pause() ;

#define FileOpen() ;

#endif

struct ios{

    template<typename ATP>ios& operator >> (ATP &x){

        x = 0; int f = 1; char c;

        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;

        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();

        x*= f;

        return *this;

    }

}io;

using namespace std;

const int N = 4000007; // how much should I open QAQ ? // Oh, I undersand ! It's influenced by 'limit'

const double pi = acos(-1.0);

struct Complex{

	double x, y;

	Complex (double xx = 0, double yy = 0) {x = xx, y = yy;}

	Complex operator + (Complex b){ return Complex(x + b.x, y + b.y); }

	Complex operator - (Complex b){ return Complex(x - b.x, y - b.y); }

	Complex operator * (Complex b){ return Complex(x * b.x - y * b.y, x * b.y + y * b.x); }

}a[N], b[N];

int r[N];

inline void FFT(int limit, Complex *a, int opt){

    R(i,0,limit - 1)

    	if(i < r[i])

    		swap(a[i], a[r[i]]);

    for(register int mid = 1; mid < limit; mid <<= 1){

    	Complex Wn( cos(pi / mid), opt * sin(pi / mid));

    	int len = mid << 1;

    	for(register int j = 0; j < limit; j += len){

    		Complex w( 1, 0);

    		R(k,0,mid - 1){

    			Complex x = a[j + k], y = w * a[j + mid + k];

    			a[j + k] = x + y;

    			a[j + mid + k] = x - y;

    			w = w * Wn;

    		}

    	}

    }

}

int main(){

	FileOpen();

	int n, m;

	io >> n >> m;

	R(i,0,n) io >> a[i].x;

	R(i,0,m) io >> b[i].x;

	int limit = 1, len = 0;

	while(limit <= n + m){

		limit <<= 1;

		++len;

	}

    R(i,0,limit - 1){

    	r[i] = (r[i >> 1] >> 1) | ((i & 1) << (len - 1));

    }

	FFT(limit, a, 1);

	FFT(limit, b, 1); // coefficient changes to point value

	R(i,0,limit){

		a[i] = a[i] * b[i];

	}

	FFT(limit, a, -1); // point value changes to coefficient

	R(i,0,n + m){

		printf("%d ", (int)(a[i].x / limit + 0.5)); // ans should divide limit

	}

	return 0;

}

巴特西

Luogu3803 【模板】多项式乘法（FFT）

递归

迭代

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