HDU 1402 FFT

题目链接

题意：

大数A*B

思路：

FFT模板题，记录一下FFT的简单应用

C++代码：

#include<bits/stdc++.h>
using namespace std;
typedef int IT;
typedef char CH;
typedef bool BL;
typedef void VD;
typedef double DB;
typedef long long LL;
typedef complex<double>CD;
const IT maxl = 131075;
const DB pi = acos(-1);
const IT maxn = 100010;

CH s1[maxl],s2[maxl];
CD a[maxl],b[maxl];
IT rev[maxl];

VD Get_rev( IT bit )
{
    for ( IT i=0 ; i<(1<<bit) ; i++ )
        rev[i] = ( rev[i>>1]>>1 )|( (i&1)<<(bit-1) );
}

VD FFT( CD *a , IT n , IT dft )
{
    for ( IT i=0 ; i<n ; i++ )
        if ( i<rev[i] )
            swap( a[i] , a[rev[i]] );
    for ( IT step=1 ; step<n ; step<<=1 )
    {
        CD wn = exp( CD( 0 , dft*pi/step ) );
        for ( IT j=0 ; j<n ; j+=step<<1 )
        {
            CD wnk( 1 , 0 );
            for ( IT k=j ; k<j+step ; k++ )
            {
                CD x = a[k];
                CD y = wnk*a[k+step];
                a[k] = x+y;
                a[k+step] = x-y;
                wnk *= wn;
            }
        }
    }
    if ( dft==-1 )
        for ( IT i=0 ; i<n ; i++ )
            a[i] /= n;
}

IT output[maxl];

IT main()
{
    for ( ; scanf ( "%s%s" , s1 , s2 )==2 ; )
    {
        IT l1 = strlen(s1);
        IT l2 = strlen(s2);
        IT s = 2,bit = 1;
        for ( ; (1<<bit)<l1+l2-1 ; bit++ )
            s <<= 1;
        for ( IT i=0 ; i<l1 ; i++ )
            a[i] = (double)(s1[l1-i-1]-'0');
        for ( IT i=l1 ; i<s ; i++ )
            a[i] = 0;
        for ( IT i=0 ; i<l2 ; i++ )
            b[i] = (double)(s2[l2-i-1]-'0');
        for ( IT i=l2 ; i<s ; i++ )
            b[i] = 0;
        Get_rev( bit );
        FFT( a , s , 1 );
        FFT( b , s , 1 );
        for ( IT i=0 ; i<s ; i++ )
            a[i] *= b[i];
        FFT( a , s , -1 );
        memset ( output , 0 , sizeof(output) );
        for ( IT i=0 ; i<s ; i++ )
        {
            output[i] += (IT)(a[i].real()+0.5);
            output[i+1] += output[i]/10;
            output[i] %= 10;
        }
        IT i;
        for ( i=l1+l2 ; !output[i]&&i>=0 ; i-- );
        if ( i==-1 )
            printf ( "0" );
        for ( ; i>=0 ; i-- )
            printf ( "%d" , output[i] );
        printf ( "\n" );
    }
    return 0;
}