hdu 1402 A * B Problem Plus FFT模板

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1402

FFT的模板题，先转成点表示法O（n）相乘之后再IDFT转回来。最后输出结果

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define mod 1000000007
#define For(i,m,n) for(int i=m;i<=n;i++)
#define Dor(i,m,n) for(int i=m;i>=n;i--)
#define LL long long
#define lan(a,b) memset(a,b,sizeof(a))
#define sqr(a) a*a
using namespace std;



const double PI = acos(-1.0);

//  复数结构体
struct Complex
{
    double x, y;    //  实部和虚部 x + yi
    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);
    }
};

//  进行FFT和IFFT前的反转变换
//  位置i和（i二进制反转后的位置）互换
//  len必须去2的幂
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]);
        }
        //  交换护卫小标反转的元素，i < j保证交换一次
        //  i做正常的+1，j左反转类型的+1，始终保持i和j是反转的
        k = len / 2;
        while (j >= k)
        {
            j -= k;
            k /= 2;
        }
        if (j < k)
        {
            j += k;
        }
    }
    return ;
}

//  FFT
//  len必须为2 ^ k形式
//  on == 1时是DFT，on == -1时是IDFT
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;
        }
    }
}
int const N=200005;
int n,m;
char a[N],b[N];
Complex x[N],y[N];
LL ans[N];
int sum[N];

int main()
{
    while(~scanf("%s%s",&a,&b))
    {
        n=strlen(a),m=strlen(b);
        int len=1;
        while(len<n*2||len<m*2)len<<=1;
        For(i,0,n-1)
        {
         x[i]=Complex(a[n-1-i]-'0',0);
        }
        For(i,n,len-1)
        {
            x[i]=Complex(0,0);
        }

        For(i,0,m-1)
        {
         y[i]=Complex(b[m-1-i]-'0',0);
        }
        For(i,m,len-1)
        {
            y[i]=Complex(0,0);
        }

        fft(x,len,1);
        fft(y,len,1);
        For(i,0,len-1)
        {
            x[i]=x[i]*y[i];
        }
        fft(x,len,-1);
        For(i,0,len-1)
            sum[i]=(int)(x[i].x+0.5);
        For(i,0,len-1)
        {
             sum[i+1]+=sum[i]/10;
            sum[i]%=10;
        }
        len=n+m-1;

        int id;
        while(sum[len]<=0&&len>0)len--;
        Dor(i,len,0)
            printf("%d",sum[i]);
        puts("");
    }

    return 0;
}