【HDU】1402 A * B Problem Plus 【FFT】
2015-04-28 11:13
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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)O(n^2)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:
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)O(n^2)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:
[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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