2016 acm香港网络赛 A.A+B problem[FFT]

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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<cstdio>#include<cmath>#include<algorithm>#include<cstring>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);}

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