Kattis aplusb A+B Problem FFT
A+B Problem
Given NN integers in the range [−50000,50000][−50000,50000], how many ways are there to pick three integers aiai, ajaj, akak, such that ii, jj, kk are pairwise distinct and ai+aj=ak? Two ways are different if their ordered triples (i,j,k)(i,j,k) of indices are different.
Input
The first line of input consists of a single integer NN (1≤N≤2000001≤N≤200000). The next line consists of NN space-separated integers a1,a2,…,aNa1,a2,…,aN.
Output
Output an integer representing the number of ways.
| Sample Input 1 | Sample Output 1 |
|---|---|
4 1 2 3 4 |
4 |
| Sample Input 2 | Sample Output 2 |
|---|---|
6 1 1 3 3 4 6 |
10 |
求 ai+aj=ak 的个数，可以转换为多项式的次数（乘法是指数相加
统计数字个数作为系数，因为有负数，所以整体+50000，注意进行多项式乘法后，变为+100000，因为是两数相加/xyx
注意0的处理：
//不同0之间相加是不应该去掉的 //不需处理
//相同数不同位置+0是不应该去掉的 //不需处理
//相同0之间的加法应该去掉 //统一去掉
//本身+0=本身应该去掉//本身也包括0的可能所以zero*(N-1) //单独去掉
细节真有趣我嘤我哭搞不清楚的时候清晰地分类列一下要
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const double PI=acos(-1.0);
const int maxn=1<<18+5;
struct Complex
{
double re,im;
Complex(double r=0.0,double i=0.0) {re=r,im=i;}
void print() {printf("%lf %lf\n",re,im);}
};
Complex operator +(const Complex&A,const Complex&B) {return Complex(A.re+B.re,A.im+B.im);}
Complex operator -(const Complex&A,const Complex&B) {return Complex(A.re-B.re,A.im-B.im);}
Complex operator *(const Complex&A,const Complex&B) {return Complex(A.re*B.re-A.im*B.im,A.re*B.im+A.im*B.re);}
Complex a[maxn],W[2][maxn];
int n,N,rev[maxn],cnt[maxn],x[maxn],z;
//bool vst[maxn];//并不需要！
LL ans;
void prework()
{
n=1<<18;
for(int i=0;i>=1,k<<=1)
(y<<=1)|=x&1;
rev[i]=y;
}
for(int i=0;i