2016 acm香港网络赛 A.A+B problem[FFT]

题意：给你一组[-50000,50000]的数，取3个数a,b,c，使得a+b=c。

思路：这题的思路是最后一小时看到FFT想到的，然后自学了一小时FFT，最后还是没搞出来。。

把每个数加上50000，这样就没有负数了，先把0拿出来，用FFT解决，再对0特殊处理一下。

#include
#include
#include
#include
using namespace std;
//kuangbin的FFT模版
const double PI = acos(-1.0);
struct complex
{
    double r,i;
    complex(double _r = 0,double _i = 0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
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]);
        k = len/2;
        while( j >= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k)j += k;
    }
}
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].r /= len;
}

const int MAXN = 800040;
complex x1[MAXN];
int a[MAXN/4];
long long num[MAXN];
long long sum[MAXN];

int main()
{
    int n=0,all,temp;
    long long num0=0; 
    scanf("%d",&all);
    memset(num,0,sizeof(num));
    memset(sum,0,sizeof(sum));
    for(int i = 0;i < all;i++)
    {
        scanf("%d",&temp);
        if(temp==0){
        	num0++;
		}else{
			a[n]=temp+50000;
			num[a[n]]++;
			sum[a[n]]++;
			n++;
		}
    }
    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++)
    	x1[i] = x1[i]*x1[i];
    fft(x1,len,-1);
    for(int i = 0;i < len;i++)
        num[i] = (long long)(x1[i].r+0.5);
    len = 2*a[n-1];
    //减去取同一个的情况
    for(int i = 0;i < n;i++)
        num[a[i]+a[i]]--;
    long long ans=0;
    int leni=a[n-1];//最大值
    for(int i = 0;i <= leni;++i){
    	if(sum[i])ans+=num[i+50000]*sum[i];//对于每个num[i],因为加上了50000,所以应该是a+b=c+50000
	}							 //乘与i的个数sum[i]就是当前的种数。
	ans+=num[100000]*num0;//a+b=0;
    if(num0!=0){
    	if(num0<3){
    		for(int i=0;i<=leni;++i){
    			if(sum[i]>=2)ans+=sum[i]*(sum[i]-1)*num0*2;//0+a=a   //0,a,a的排列方式有C(2,1)*C(2,1) 再乘上 C(num0,1),C(sum[i],2) 
			}
		}else if(num0>=3){
			for(int i=0;i<=leni;++i){
    			if(sum[i]>=2)ans+=sum[i]*(sum[i]-1)*num0*2;
			}
			ans+=num0*(num0-1)*(num0-2);//0+0=0    //C(num0,3)*A(3,3)
		}
	}
	printf("%lld\n",ans);
}