【FFT】 SPOJ Triple Sums

先類似計數排序一樣求一下數的個數，求一下p的3次方的卷積，然後類似容斥一樣減去重複的就可以了。。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 300005
#define maxm 1000005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

struct complex
{
	double r, i;
	complex(double r = 0, double i = 0) : r(r), i(i) {}
	complex operator + (complex b) const {
		return complex(r + b.r, i + b.i);
	}
	complex operator - (complex b) const {
		return complex(r - b.r, i - b.i);
	}
	complex operator * (complex b) const {
		return complex(r * b.r - i * b.i, r * b.i + i * b.r);
	}
}A[maxn], B[maxn];

void fft(complex y[], int len, int on)
{
	for(int i = 1, j = len / 2; i < len-1; i++) {
		if(i < j) swap(y[i], y[j]);
		int k = len / 2;
		while(j >= k) {
			j -= k;
			k /= 2;
		}
		j += k;
	}
	for(int i = 2; i <= len; i <<= 1) {
		complex wn(cos(-on * 2 * PI / i), sin(-on * 2 * PI / i));
		for(int j = 0; j < len; j += i) {
			complex w(1, 0);
			for(int k = j; k < j + i / 2; k++) {
				complex u = y[k];
				complex t = y[k + i / 2] * w;
				y[k] = u + t;
				y[k + i / 2] = u - t;
				w = w * wn;
			}
		}
	}
	if(on == -1) for(int i = 0; i < len; i++) y[i].r /= len;
}  

const int base = 20000;
const complex three = complex(3.0, 0);
int a[maxn];
int b[maxn];
int c[maxn];
int mx, n;

int cmp(int a, int b)
{
	return a < b;
}

void init(void)
{
	memset(a, 0, sizeof a);
	memset(b, 0, sizeof b);
	memset(c, 0, sizeof c);
}

void read(void)
{
	int x;
	mx = -INF;
	for(int i = 0; i < n; i++) {
		scanf("%d", &x);
		x += base;
		mx = max(mx, x);
		a[x]++;
		b[x + x]++;
		c[x + x + x]++;
	}
}

void work(void)
{
	int len1 = mx+1, len = 1;
	while(len < 4 * len1) len <<= 1;
	for(int i = 0; i < len; i++) A[i] = complex(a[i], 0);
	for(int i = 0; i < len; i++) B[i] = complex(b[i], 0);
	fft(A, len, 1);
	fft(B, len, 1);
	for(int i = 0; i < len; i++) A[i] = A[i] * (A[i] * A[i] - three * B[i]);
	fft(A, len, -1);
	for(int i = 0; i < len; i++) {
		int ans = ((int)(A[i].r + 0.5) + 2 * c[i]) / 6;
		if(ans) printf("%d : %d\n", i - base * 3, ans);
	}
}

int main(void)
{
	while(scanf("%d", &n) != EOF) {
		init();
		read();
		work();
	}

	return 0;
}