提交记录 3204


用户 题目 状态 得分 用时 内存 语言 代码长度
wys 1004a. 【模板题】高精度乘法2 Accepted 100 2.015 ms 8428 KB C++ 5.24 KB
提交时间 评测时间
2018-07-10 14:53:29 2020-07-31 21:12:33
#include <stdio.h>
#include <string.h>
#include <math.h>

#define debug(...) fprintf(stderr, __VA_ARGS__)

#define PI 3.14159265358979323846

template <class T>
inline void swap(T &a, T &b) {
	T t = a;
	a = b, b = t;
}

namespace IO {
	const int BUF_SIZE = 1 << 15;
	char in_buf[BUF_SIZE], out_buf[BUF_SIZE];
	char *p_in_buf = in_buf + BUF_SIZE;
	char *p_out_buf = out_buf;
	
	
	inline char get_char() {
		if (p_in_buf == in_buf + BUF_SIZE) {
			fread(in_buf, 1, BUF_SIZE, stdin), p_in_buf = in_buf;
		}
		return *(p_in_buf++);
	}
	
	inline void put_char(char x) {
		if (p_out_buf == out_buf + BUF_SIZE) {
			fwrite(out_buf, 1, BUF_SIZE, stdout), p_out_buf = out_buf;
		}
		*(p_out_buf++) = x;
	}
	
	inline void flush() {
		if (p_out_buf != out_buf) {
			fwrite(out_buf, 1, p_out_buf - out_buf, stdout);
			p_out_buf = out_buf;
		}
	}
}

#define getchar() IO::get_char()
#define putchar(x) IO::put_char(x)

inline int getint() {
	int x = 0;
	char c = getchar();
	while (c <= 32) c = getchar();
	while (c > 32) x = x * 10 + c - 48, c = getchar();
	return x;
}

inline int getint(char &c) {
	int x = 0;
	c = getchar();
	while (c <= 32) c = getchar();
	while (c > 32) x = x * 10 + c - 48, c = getchar();
	return x;
}

template <class T>
inline void _putint(T x) {
	return x ? _putint(x / 10), putchar(48 + x % 10), void() : void();
}

template <class T>
inline void putint(T x) {
	return x == 0 ? putchar('0'), void() : _putint(x), void();
}

inline void getline(char *s) {
	char c = getchar();
	while (c == '\n') c = getchar();
	while (c != '\n') *(s++) = c, c = getchar();
	*s = 0;
}

// ==== header ====

#define float double

struct Complex {
	float x, y;
	
	Complex conj() {
		return (Complex){x, -y};
	}
	
	Complex conj2() {
		return (Complex){-x, y};
	}
};

inline Complex operator + (const Complex &a, const Complex &b) {
	return (Complex){a.x + b.x, a.y + b.y};
}

inline Complex operator - (const Complex &a, const Complex &b) {
	return (Complex){a.x - b.x, a.y - b.y};
}

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

Complex A[(1 << 21) + 1], B[(1 << 21) + 1];

int bitrev[1 << 11];

void bitrev_init() {
	for (int i = 0; i < 1 << 11; i++) {
		int t = 0;
		for (int j = 0; j < 11; j++) {
			t |= ((i >> j) & 1) << (10 - j);
		}
		bitrev[i] = t;
	}
}

inline int get_bitrev(int x, int len) {
	return (bitrev[x >> 11] | (bitrev[x & ((1 << 11) - 1)] << 11)) >> (22 - len);
}

void FFT(Complex *a, int lg_n, bool rev) {
	int n = 1 << lg_n;
	
	for (int i = lg_n - 1; i >= 0; i--) {
		int S = 1 << i;
		Complex w1 = (Complex){cos(PI / S), -sin(PI / S) * (rev ? -1 : 1)};
		for (int j = 0; j < n; j += 2 * S) {
			Complex w = (Complex){1.0, 0.0};
			Complex *A = a + j;
			for (int k = 0; k < S; k++) {
				Complex t = A[k + S];
				A[k + S] = (A[k] - t) * w;
				A[k] = A[k] + t;
				w = w * w1;
			}
		}
	}
	
	for (int i = 0; i < n; i++) {
		int t = get_bitrev(i, lg_n);
		if (i < t) {
			Complex tmp = a[i];
			a[i] = a[t], a[t] = tmp;
		}
	}
}

void FFT(Complex *a, Complex *b, int lg_n, bool rev) {
	int n = 1 << lg_n;
	
	for (int i = lg_n - 1; i >= 0; i--) {
		int S = 1 << i;
		Complex w1 = (Complex){cos(PI / S), -sin(PI / S) * (rev ? -1 : 1)};
		for (int j = 0; j < n; j += 2 * S) {
			Complex w = (Complex){1.0, 0.0};
			Complex *A = a + j, *B = b + j;
			for (int k = 0; k < S; k++) {
				Complex ta = A[k + S], tb = B[k + S];
				A[k + S] = (A[k] - ta) * w;
				B[k + S] = (B[k] - tb) * w;
				A[k] = A[k] + ta;
				B[k] = B[k] + tb;
				w = w * w1;
			}
		}
	}
	
	for (int i = 0; i < n; i++) {
		int t = get_bitrev(i, lg_n);
		if (i < t) {
			Complex tmp = a[i];
			a[i] = a[t], a[t] = tmp;
			tmp = b[i];
			b[i] = b[t], b[t] = tmp;
		}
	}
}

char in_s[1 << 18];

int main() {
	int n, m;
	n = m = 10000 - 1;
	
	int lg_n = 3;
	while (1 << lg_n < n + 1 || 1 << lg_n < m + 1) ++lg_n;
	
	bitrev_init();
	
	int N = 1 << lg_n;
	int N2 = N >> 1;
	
	getline(in_s);
	for (int i = 0; i <= n; i++) {
		// *0.5 is to avoid *2 in later caclulation
		(i & 1 ? A[i >> 1].y : A[i >> 1].x) = (in_s[n - i] - 48) * 0.5;
	}
	
	getline(in_s);
	for (int i = 0; i <= m; i++) {
		(i & 1 ? B[i >> 1].y : B[i >> 1].x) = (in_s[m - i] - 48) * 0.5;
	}
	
	FFT(A, B, lg_n, false);
	
	A[N] = A[0];
	B[N] = B[0];
	const Complex w = (Complex){cos(2 * PI / N), sin(2 * PI / N)};
	Complex w_product = (Complex){1, 0};
	for (int i = 0; i <= N2; i++) {
		Complex a1 = A[i] + A[N - i];
		Complex a2 = A[i] - A[N - i];
		Complex b1 = B[i] + B[N - i];
		Complex b2 = B[i] - B[N - i];
		
		Complex a = (Complex){a1.x, a2.y};
		Complex b = (Complex){a2.x, a1.y};
		Complex c = (Complex){b1.x, b2.y};
		Complex d = (Complex){b2.x, b1.y};
		
		// a * c - b * d * delta(x - 1) + a * d + b * c
		// @Signal Processing
		A[i] = a * c + a * d + b * c - b * d * w_product.conj();
		A[N - i] = (a * c).conj() + a.conj() * d.conj2() + b.conj2() * c.conj() - b.conj2() * d.conj2() * w_product;
		w_product = w_product * w;
	}
	
	FFT(A, lg_n, true);
	
	float inv_n = 1.0 / N;
	int ans[2000005];
	for (int i = 0; i <= n + m; i++) {
		ans[i] = (int)((i & 1 ? A[i >> 1].y : A[i >> 1].x) * inv_n + 0.5);
	}
	ans[n + m + 1] = 0;
	for (int i = 0; i <= n + m; i++) {
		ans[i + 1] += ans[i] / 10;
		ans[i] = ans[i] % 10;
	}
	int i = n + m + 1;
	while (!ans[i]) i--;
	while (i >= 0) putchar(ans[i] + 48), i--;
	
	IO::flush();
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #12.015 ms8 MB + 236 KBAcceptedScore: 100


Judge Duck Online | 评测鸭在线
Server Time: 2024-10-05 12:27:33 | Loaded in 6 ms | Server Status
个人娱乐项目,仅供学习交流使用