邝斌的ACM模板（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int
long 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;
}