FFT模板

#include <cstdio>

#include <cmath>

#include <cstring>

using namespace std;

const double pi = acos(-1.0);

const int maxn = 500005;

struct complex

{

    double r,i;

    complex(double r = 0.0,double i = 0.0) :r(r),i(i) {}

    inline complex operator + (const complex &b)   const

    {

        return complex(r + b.r,i + b.i);

    }

    inline complex operator - (const complex &b)   const

    {

        return complex(r - b.r,i - b.i);

    }

    inline complex operator * (const complex &b)   const

    {

        return complex(r * b.r - i * b.i,r * b.i + i * b.r);

    }

    friend inline void swap(complex &a,complex &b)  {   complex t = a;a = b;b = t;  }

};

complex a[maxn << 1],b[maxn << 1];

int ans[maxn << 1];

char szA[maxn],szB[maxn];

inline void fft(complex * y,int n,int on)

{

    for(int mid = n >> 1,i = 1,j = mid;i < n - 1;i ++)

    {

        if(i < j)   swap(y[i],y[j]);    int k = mid;

        for(;j >= k;k >>= 1) j -= k; j += k;

    }

    for(int h = 2;h <= n;h <<= 1)

    {

        complex wn(cos(2 * pi / h),sin(on * 2 * pi / h));

        for(int j = 0;j < n;j += h)

        {

            complex w(1,0); int midh = h >> 1;

            for(int k = j;k < j + midh;k ++,w = w * wn)

            {

                complex u = y[k],t = w * y[k + midh];

                y[k] = u + t;   y[k + midh] = u - t;

            }

        }

    }

    if(on == -1)    for(int i = 0;i < n;i ++)   y[i].r /= n;

}

int main()

{

    int lenA,lenB,len;

    while(~scanf("%s%s",szA,szB))

    {

        lenA = strlen(szA),lenB = strlen(szB);

        for(len = 1;len < (lenA << 1) || len < (lenB << 1);len <<= 1);

        for(int i = 0;i < lenA;i ++)    a[i].r = szA[lenA - i - 1] - ‘0‘,a[i].i = 0;

        for(int i = lenA;i < len;i ++)  a[i].r = a[i].i = 0;

        for(int i = 0;i < lenB;i ++)    b[i].r = szB[lenB - i - 1] - ‘0‘,b[i].i = 0;

        for(int i = lenB;i < len;i ++)  b[i].r = b[i].i = 0;

        fft(a,len,1);   fft(b,len,1);

        for(int i = 0;i < len;i ++) a[i] = a[i] * b[i];

        fft(a,len,-1);

        for(int i = 0;i < len;i ++)

            ans[i] = a[i].r + 0.5;

        for(int i = 0;i < len;i ++) if(ans[i] > 9)  ans[i + 1] += ans[i] / 10,ans[i] %= 10;

        int p = lenA + lenB - 1;

        while(!ans[p] && p) p --;

        while(p >= 0)   putchar(ans[p --] + ‘0‘);

        putchar(‘\n‘);

    }

    return 0;

}

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