HDU 1402 FFT模板题，求大数乘法

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这两天刚开始看FFT，有点云里雾里的，今天看了个模板题，觉得这位聚聚写得特别好，总算是看懂了多项式相乘之类的运算，存一下。

http://blog.csdn.net/sdj222555/article/details/9786527

好点复杂一点的算法，数据结构都不知道，还要慢慢学习。

FFT模板：

#define L(x) (1 << (x))const double PI = acos(-1.0);const int Maxn = 133015;double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];int revv(int x, int bits){    int ret = 0;    for (int i = 0; i < bits; i++)    {        ret <<= 1;        ret |= x & 1;        x >>= 1;    }    return ret;}void fft(double * a, double * b, int n, bool rev){    int bits = 0;    while (1 << bits < n) ++bits;    for (int i = 0; i < n; i++)    {        int j = revv(i, bits);        if (i < j)            swap(a[i], a[j]), swap(b[i], b[j]);    }    for (int len = 2; len <= n; len <<= 1)    {        int half = len >> 1;        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);        if (rev) wmy = -wmy;        for (int i = 0; i < n; i += len)        {            double wx = 1, wy = 0;            for (int j = 0; j < half; j++)            {                double cx = a[i + j], cy = b[i + j];                double dx = a[i + j + half], dy = b[i + j + half];                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;                a[i + j] = cx + ex, b[i + j] = cy + ey;                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;                wx = wnx, wy = wny;            }        }    }    if (rev)    {        for (int i = 0; i < n; i++)            a[i] /= n, b[i] /= n;    }}int solve(int a[],int na,int b[],int nb,int ans[]) //两个数组求卷积,有时ans数组要开成long long{    int len = max(na, nb), ln;    for(ln=0; L(ln)<len; ++ln);    len=L(++ln);    for (int i = 0; i < len ; ++i)    {        if (i >= na) ax[i] = 0, ay[i] =0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for (int i = 0; i < len; ++i)    {        if (i >= nb) bx[i] = 0, by[i] = 0;        else bx[i] = b[i], by[i] = 0;    }    fft(bx, by, len, 0);    for (int i = 0; i < len; ++i)    {        double cx = ax[i] * bx[i] - ay[i] * by[i];        double cy = ax[i] * by[i] + ay[i] * bx[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for (int i = 0; i < len; ++i)        ans[i] = (int)(ax[i] + 0.5);    return len;}int solve(long long a[], int na, int ans[]) //自己跟自己求卷积,有时候ans数组要开成long long{    int len = na, ln;    for(ln = 0; L(ln) < na; ++ln);    len=L(++ln);    for(int i = 0; i < len; ++i)    {        if (i >= na) ax[i] = 0, ay[i] = 0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for(int i=0; i<len; ++i)    {        double cx = ax[i] * ax[i] - ay[i] * ay[i];        double cy = 2 * ax[i] * ay[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for(int i=0; i<len; ++i)        ans[i] = ax[i] + 0.5;    return len;}

大数相乘模板（按照FFT的规则，低位要放在前面，其实多项式相乘就是多项式系数的相乘，而多项式系数的相乘跟我们这里的大数相乘就是一样的东西了，最后再倒序输出就好）

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <queue>#include <set>#include <vector>using namespace std;#define L(x) (1 << (x))const double PI = acos(-1.0);const int Maxn = 133015;double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];char sa[Maxn/2],sb[Maxn/2];int sum[Maxn];int x1[Maxn],x2[Maxn];int revv(int x, int bits){    int ret = 0;    for (int i = 0; i < bits; i++)    {        ret <<= 1;        ret |= x & 1;        x >>= 1;    }    return ret;}void fft(double * a, double * b, int n, bool rev){    int bits = 0;    while (1 << bits < n) ++bits;    for (int i = 0; i < n; i++)    {        int j = revv(i, bits);        if (i < j)            swap(a[i], a[j]), swap(b[i], b[j]);    }    for (int len = 2; len <= n; len <<= 1)    {        int half = len >> 1;        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);        if (rev) wmy = -wmy;        for (int i = 0; i < n; i += len)        {            double wx = 1, wy = 0;            for (int j = 0; j < half; j++)            {                double cx = a[i + j], cy = b[i + j];                double dx = a[i + j + half], dy = b[i + j + half];                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;                a[i + j] = cx + ex, b[i + j] = cy + ey;                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;                wx = wnx, wy = wny;            }        }    }    if (rev)    {        for (int i = 0; i < n; i++)            a[i] /= n, b[i] /= n;    }}int solve(int a[],int na,int b[],int nb,int ans[]){    int len = max(na, nb), ln;    for(ln=0; L(ln)<len; ++ln);    len=L(++ln);    for (int i = 0; i < len ; ++i)    {        if (i >= na) ax[i] = 0, ay[i] =0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for (int i = 0; i < len; ++i)    {        if (i >= nb) bx[i] = 0, by[i] = 0;        else bx[i] = b[i], by[i] = 0;    }    fft(bx, by, len, 0);    for (int i = 0; i < len; ++i)    {        double cx = ax[i] * bx[i] - ay[i] * by[i];        double cy = ax[i] * by[i] + ay[i] * bx[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for (int i = 0; i < len; ++i)        ans[i] = (int)(ax[i] + 0.5);    return len;}int main(){    int l1,l2,l;    int i;    while(gets(sa))    {        gets(sb);        memset(sum, 0, sizeof(sum));        l1 = strlen(sa);        l2 = strlen(sb);        for(i = 0; i < l1; i++)            x1[i] = sa[l1 - i - 1]-'0';        for(i = 0; i < l2; i++)            x2[i] = sb[l2-i-1]-'0';        l = solve(x1, l1, x2, l2, sum);        for(i = 0; i<l || sum[i] >= 10; i++) // 进位        {            sum[i + 1] += sum[i] / 10;            sum[i] %= 10;        }        l = i;        while(sum[l] <= 0 && l>0)    l--; // 检索最高位        for(i = l; i >= 0; i--)    putchar(sum[i] + '0'); // 倒序输出        putchar('\n');    }    return 0;}

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