邝斌的ACM模板（FFT）

本博客整理自邝斌的ACM模板
2.9、FFT

//HDU 1402 求高精度乘法#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <math.h>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]);        //交换互为小标反转的元素，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;    }}/* * 做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;}const int MAXN = 200010;Complex x1[MAXN],x2[MAXN];char str1[MAXN/2],str2[MAXN/2];int sum[MAXN];int main(){    while(scanf("%s%s",str1,str2)==2)    {        int len1 = strlen(str1);        int len2 = strlen(str2);        int len = 1;        while(len < len1*2 || len < len2*2)len<<=1;        for(int i = 0; i < len1; i++)            x1[i] = Complex(str1[len1-1-i]-'0',0);        for(int i = len1; i < len; i++)            x1[i] = Complex(0,0);        for(int i = 0; i < len2; i++)            x2[i] = Complex(str2[len2-1-i]-'0',0);        for(int i = len2; i < len; i++)            x2[i] = Complex(0,0);        //求DFT        fft(x1,len,1);        fft(x2,len,1);        for(int i = 0; i < len; i++)            x1[i] = x1[i]*x2[i];        fft(x1,len,-1);        for(int i = 0; i < len; i++)            sum[i] = (int)(x1[i].x+0.5);        for(int i = 0; i < len; i++)        {            sum[i+1]+=sum[i]/10;            sum[i]%=10;        }        len = len1+len2-1;        while(sum[len] <= 0 && len > 0)len--;        for(int i = len; i >= 0; i--)            printf("%c",sum[i]+'0');        printf("\n");    }    return 0;}

//HDU 4609//给出 n 条线段长度，问任取 3 根，组成三角形的概率。//n<=10^5   用 FFT 求可以组成三角形的取法有几种const int MAXN = 400040;Complex x1[MAXN];int a[MAXN/4];long long num[MAXN];//100000*100000会超intlong long sum[MAXN];int main(){    int T;    int n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        memset(num,0,sizeof(num));        for(int i = 0; i < n; i++)        {            scanf("%d",&a[i]);            num[a[i]]++;        }        sort(a,a+n);        int len1 = a[n-1]+1;        int len = 1;        while( len < 2*len1 )len <<= 1;        for(int i = 0; i < len1; i++)            x1[i] = Complex(num[i],0);        for(int i = len1; i < len; i++)            x1[i] = Complex(0,0);        fft(x1,len,1);        for(int i = 0; i < len; i++)             x            1[i] = x1[i]*x1[i];        fft(x1,len,-1);        for(int i = 0; i < len; i++)            num[i] = (long long)(x1[i].x+0.5);        len = 2*a[n-1];        //减掉取两个相同的组合        for(int i = 0; i < n; i++)            num[a[i]+a[i]]--;        for(int i = 1; i <= len; i++)num[i]/=2;        sum[0] = 0;        for(int i = 1; i <= len; i++)            sum[i] = sum[i-1]+num[i];        long long cnt = 0;        for(int i = 0; i < n; i++)        {            cnt += sum[len]-sum[a[i]];            //减掉一个取大，一个取小的            cnt -= (long long)(n-1-i)*i;            //减掉一个取本身，另外一个取其它            cnt -= (n-1);            cnt -= (long long)(n-1-i)*(n-i-2)/2;        }        long long tot = (long long)n*(n-1)*(n-2)/6;        printf("%.7lf\n",(double)cnt/tot);    }    return 0;}