SPOJ：Triple Sums（母函数+FFT）

You're given a sequence s of N distinct integers.
Consider all the possible sums of three integers from the sequence at three different indicies.
For each obtainable sum output the number of different triples of indicies that generate it.

Constraints:
N <= 40000, |s_i| <= 20000

Input

The first line of input contains a single integer N.
Each of the next N lines contain an element of s.

Output

Print the solution for each possible sum in the following format:
sum_value : number_of_triples

Smaller sum values should be printed first.

Example

Input:

5
-1
2
3
0
5
Output:

1 : 1
2 : 1
4 : 2
5 : 1
6 : 1
7 : 2
8 : 1
10 : 1

Explanation:
4 can be obtained using triples ( 0, 1, 2 ) and ( 0, 3, 4 ).
7 can be obtained using triples ( 0, 2, 4 ) and ( 1, 3, 4 ).

Note: a triple is considered the same as any of its permutations.

2018/4/4，因为前几天一直在做LCT，然后基础题都长得差不多，难一点的现在还没有相同，然后搞得有些烦躁。就干脆转移下注意力。

于是学习了下FFT，算是以为数学大渣又跨越了一小步。

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这题可以先用母函数表示出选一个的方案（系数是物品出现次数，指数是物品价值）

A(x)=a*x^1+b*x^2+c*x^3

所以不考虑重复，在物品中选出三个的方案就是 1/6*A(x)^3
现在用 B(x),C(x) 分别表示一种物品选了 2 次和 3 次的方案

B(x)=a*x^2+b*x^4+c*x^6  C(x)=a*x^3+b*x^6+c*x^9


选三个有可能是 AAB, ABA, BAA, AAA 这几种重复情况，所以扣掉后方案就是

[A(x)^3−3*A(x)⋅B(x)+2*C(x)]/6

由于这里用到了多项式乘法，用FFT优化即可

由于x的范围可能为负，所以假设每个数都加20000，使之变为正，最后再减去即可。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
const int maxn=1<<17;
const double pi=acos(-1.0);
using namespace std;
struct complex{
    double r,i;
    complex(){};
    complex(double rr,double ii):r(rr),i(ii){}
    complex friend operator +(complex a,complex b){return (complex){a.r+b.r,a.i+b.i};}
    complex friend operator -(complex a,complex b){return (complex){a.r-b.r,a.i-b.i};}
    complex friend operator *(complex a,complex b){return (complex){a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r};}
}tmp[maxn];
struct DFT{
    complex a[maxn];
    void fft(int sz,int bg,int step,int opt){
       if(sz==1) return;  int m=sz>>1;
       fft(m,bg,step<<1,opt); fft(m,bg+step,step<<1,opt);
       complex w=complex(1,0),t=complex(cos(2.0*pi/sz),sin(2.0*pi*opt/sz)); 
       for(int k=0;k<m;k++)
       {
            int pos=2*step*k;
          tmp[k]=a[pos+bg]+w*a[pos+bg+step];
          tmp[k+m]=a[pos+bg]-w*a[pos+bg+step];
          w=w*t;
       }
       for(int i=0;i!=sz;i++) a[i*step+bg]=tmp[i];
    }
}A,B,C;
int a[maxn],b[maxn],c[maxn];
int main(){
    int n; scanf("%d",&n);
    for(int i=0,x;i<n;i++) scanf("%d",&x),a[x+20000]++,b[2*(x+20000)]++,c[3*(x+20000)]++;
    for(int i=0;i<maxn;i++) A.a[i].r=a[i], B.a[i].r=b[i], C.a[i].r=c[i];
    A.fft(maxn,0,1,1), B.fft(maxn,0,1,1);
    for(int i=0;i<maxn;i++) C.a[i]=A.a[i]*(A.a[i]*A.a[i]-(complex){3.0,0.0}*B.a[i]);
    C.fft(maxn,0,1,-1);
    for(int i=0;i<maxn;i++){
        ll ans=((ll)(C.a[i].r/maxn+0.5)+2*c[i])/6;
        if (ans) printf("%d : %I64d
",i-60000,ans);
    }
    return 0;
    return 0;
}