Asia Hong Kong Online Preliminary A (FFT)

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题目链接：点击这里

题意：给出n个数字（∈[−50000,50000]）,求有多少(i,j,k)满足ai+aj=ak(要求i,j,k互不相同)。

0比较麻烦直接拎出来记录个数，其他的加一个50000避免负数直接存起来。然后开一个数组记录每一个数字出现多少次，这个数组和自己做一次卷积即可。答案分成几部分：
1. i+j=k: 直接遍历扔掉0的数组，把算完卷积的结果扔进去；
2. 0+0=0：直接用0的个数可以得出结果；
3. x+y=0：直接用卷积中0次项的系数或者暴力遍历；
4. 0+x=x/x+0=x：暴力遍历统计。

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <map>using namespace std;#define pi acos (-1)  #define maxn 511111#define add 50000struct plex {      double x, y;      plex (double _x = 0.0, double _y = 0.0) : x (_x), y (_y) {}      plex operator + (const plex &a) const {          return plex (x+a.x, y+a.y);      }      plex operator - (const plex &a) const {          return plex (x-a.x, y-a.y);      }      plex operator * (const plex &a) const {          return plex (x*a.x-y*a.y, x*a.y+y*a.x);      }  }x[maxn];  void change (plex *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(plex y[],int len,int on)  {      change(y,len);      for(int h = 2; h <= len; h <<= 1)      {          plex wn(cos(-on*2*pi/h),sin(-on*2*pi/h));          for(int j = 0;j < len;j+=h)          {              plex w(1,0);              for(int k = j;k < j+h/2;k++)              {                  plex u = y[k];                  plex 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;  }  int n, cnt, zero;int a[maxn], num[maxn];long long solve () {    long long ans = 0;    int len = 1;    while (len < 200000) {        len <<= 1;    }    for (int i = 0; i < len; i++) {        x[i] = plex (num[i], 0);    }    fft (x, len, 1);    for (int i = 0; i < len; i++) {        x[i] = x[i]*x[i];    }    fft (x, len, -1);    for (int i = 1; i <= cnt; i++) {//i+j=k        long long tmp = (long long) (x[a[i]+add*2].x + 0.5);        ans += tmp;    }    for (int i = 1; i <= cnt; i++) {//减去i+i=j        if (a[i]%2 == 0)            ans -= num[a[i]/2+add];    }    if (zero >= 3) ans += 1LL*zero*(zero-1)*(zero-2);//0+0=0    for (int i = 0; i <= add*2; i++) {//x+0=y y+0=x        if (num[i] > 1) {            ans += 1LL*2*num[i]*(num[i]-1)*zero;        }    }    ans += 1LL*(long long)(x[2*add].x+0.5)*zero;//x+y=0    return ans;}int main () {    scanf ("%d", &n);    zero = cnt = 0;    memset (num, 0, sizeof num);    for (int i = 1; i <= n; i++) {        scanf ("%d", &a[++cnt]);        if (a[cnt] == 0) {            zero++;            cnt--;            continue;        }        num[a[cnt]+add]++;    }    long long ans = solve ();    printf ("%lld\n", ans);    return 0;}/*10-50000 -50000 -50000 0 0 0 50000 50000 50000 140 0 -1 1*/

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