【HDU】1402 A * B Problem Plus 【FFT】

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传送门：【HDU】1402 A * B Problem Plus

题目分析：

这就是大数乘法题，问两个大数相乘的结果，由于O(n2)的算法复杂度太大，所以我们用FFT来优化他。关于FFT网上资料很多，我就不多说啦。

这是我做的第一道FFT，FFT是看算法导论学来的，感觉算导讲的很不错，简单易懂~

my code：

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 200005 ;const double pi = acos ( -1.0 ) ;struct Complex {    double r , i ;    Complex () {}    Complex ( double r , double i ) : r ( r ) , i ( i ) {}    Complex operator + ( const Complex& t ) const {        return Complex ( r + t.r , i + t.i ) ;    }    Complex operator - ( const Complex& t ) const {        return Complex ( r - t.r , i - t.i ) ;    }    Complex operator * ( const Complex& t ) const {        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;    }} ;void FFT ( Complex y[] , int n , int rev ) {    for ( int i = 1 , j , k , t ; i < n ; ++ i ) {        for ( j = 0 , k = n >> 1 , t = i ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;        if ( i < j ) swap ( y[i] , y[j] ) ;    }    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w = Complex ( 1 , 0 ) , t ;        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {            for ( int i = k ; i < n ; i += s ) {                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;                y[i] = y[i] + t ;            }        }    }    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;}char s1[MAXN] , s2[MAXN] ;Complex x1[MAXN] , x2[MAXN] ;int num[MAXN] ;void solve () {    int n1 = strlen ( s1 ) ;    int n2 = strlen ( s2 ) ;    int n = 1 ;    while ( n < n1 + n2 ) n <<= 1 ;    for ( int i = 0 ; i < n1 ; ++ i ) x1[i] = Complex ( s1[n1 - i - 1] - '0' , 0 ) ;    for ( int i = n1 ; i < n ; ++ i ) x1[i] = Complex ( 0 , 0 ) ;    for ( int i = 0 ; i < n2 ; ++ i ) x2[i] = Complex ( s2[n2 - i - 1] - '0' , 0 ) ;    for ( int i = n2 ; i < n ; ++ i ) x2[i] = Complex ( 0 , 0 ) ;    FFT ( x1 , n , 1 ) ;    FFT ( x2 , n , 1 ) ;    for ( int i = 0 ; i < n ; ++ i ) x1[i] = x1[i] * x2[i] ;    FFT ( x1 , n , -1 ) ;    int t = 0 ;    for ( int i = 0 ; i < n ; ++ i , t /= 10 ) {        t += ( int ) ( x1[i].r + 0.1 ) ;        num[i] = t % 10 ;    }    for ( ; t ; t /= 10 ) num[n ++] = t % 10 ;    while ( n > 1 && !num[n - 1] ) -- n ;    for ( int i = n - 1 ; i >= 0 ; -- i ) printf ( "%d" , num[i] ) ;    printf ( "\n" ) ;}int main () {    while ( ~scanf ( "%s%s" , s1 , s2 ) ) solve () ;    return 0 ;}

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