Luogu1919 【模板】A*B Problem升级版（FFT）

简单的\(A*B\) \(Problem\)，卡精度卡到想女装

#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 type = 1; char c;

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

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

        x*= type;

        return *this;

    }

}io;

using namespace std;

const int N = 240000; // 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 type){

    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), type * 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 c[N];

int main(){

//	FileOpen();

	int n;

    io >> n;

	--n;

	int m = n << 1;

	R(i,0,n) scanf("%1lf", &a[n - i].x);

	R(i,0,n) scanf("%1lf", &b[n - i].x);

    int limit = 1, len = 0;

    while(limit <= m){

    	++len;

    	limit <<= 1;

    }

    R(i,0,limit - 1){

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

    }

    FFT(limit, a, 1);

	FFT(limit, b, 1);

    R(i,0,limit){

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

    }

    FFT(limit, a, -1);

    R(i,0,limit)

        a[i].x /= limit;

    R(i,0,m){

    	c[i] = (int)(a[i].x + 0.1);

    }

    R(i,0,m)

        if(c[i] > 9){

        	c[i + 1] += c[i] / 10;

        	c[i] %= 10;

            if(i == m)

				++m;

        }

    while(c[m] == 0) --m;

    nR(i,m,0){

		printf("%d", c[i]);

    }

    return 0;

}

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