Luogu3803 【模板】多项式乘法（FFT）

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;
}