HDU 1402 A * B fft 板子题
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1402
题目描述: 计算A * B, 其中A, B是长度小于等于50000的整数
解题思路: 放在以前我可能会拿两个高精度的乘法来计算, 但现在我接触到了fft: 求值 -> 相乘 -> 插值。下面是一篇很好的 fft 的讲解:
http://www.gatevin.moe/acm/fft%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0/
首先我们把A, B分解成系数表达式, 也就是十进制表示的所有系数前面的整数, 如12345就是(1,2,3,4,5), 然后fft, 板子题。
题目代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const double PI = acos(-1.0); struct Complex { double x, y; Complex(double _x = 0.0, double _y = 0.0) { x = _x; y = _y; } Complex operator -(const Complex & b )const { return Complex(x-b.x, y-b.y); } Complex operator +(const Complex & b)const { return Complex(x+b.x, y+b.y); } Complex operator *(const Complex & b)const { return Complex(x*b.x-y*b.y, x*b.y+y*b.x); } }; void change(Complex y[], int len) { int i, j, k; for(i = 1, j = len / 2; i < len - 1; i++ ) { if( i < j ) swap( y[i], y[j] ); k = len / 2; while( j >= k ) { j -= k; k /= 2; } if( j < k ) j += k; } } void fft(Complex y[], int len, int on) { change( y, len ); for( int h = 2; h <= len; h <<= 1 ) { Complex wn( cos(-on*2*PI/h), sin(-on*2*PI/h) ); for( int j = 0; j < len; j += h) { Complex w(1, 0); for( int k = j; k < j + h / 2; k++ ) { Complex u = y[k]; Complex t = w * y[k+h/2]; y[k] = u + t; y[k+h/2] = u - t; w = w * wn; } } } if( on == -1 ) { for( int i = 0; i < len; i++ ) { y[i].x /= len; } } } const int MAXN = 200010; Complex x1[MAXN], x2[MAXN]; char str1[MAXN/2], str2[MAXN/2]; int sum[MAXN]; int main() { while( scanf( "%s%s", str1, str2 ) == 2 ) { int len1 = (int)strlen(str1); int len2 = (int)strlen(str2); int len = 1; while( len < len1 * 2 || len < len2 * 2 ) len <<= 1; for( int i = 0; i < len1; i++ ) { x1[i] = Complex(str1[len1-1-i]-'0', 0); } for( int i = len1; i < len; i++ ) { x1[i] = Complex(0,0); } for( int i = 0; i < len2; i++ ) { x2[i] = Complex(str2[len2-1-i]-'0', 0); } for( int i = len2; i < len; i++ ) { x2[i] = Complex(0,0); } fft( x1, len, 1 ); fft( x2, len, 1 ); for( int i = 0; i < len; i++ ) { x1[i] = x1[i] * x2[i]; } fft( x1, len, -1 ); for( int i = 0; i < len; i++ ) { sum[i] = (int)(x1[i].x + 0.5); } for( int i = 0; i < len; i++ ) { sum[i+1] += sum[i] / 10; sum[i] %= 10; } len = len1 + len2 -1; while( sum[len] <= 0 && len > 0 ) len--; for( int i = len; i >= 0; i-- ) { printf( "%c", sum[i] + '0' ); } printf( "\n" ); } return 0; }
思考: 多做数学题, fft, 高斯消元, 欧几里得定理等等。
posted on 2017-05-03 17:07 FriskyPuppy 阅读(213) 评论(0) 编辑 收藏 举报