[SPOJ TSUM]Triple Sums(FFT+容斥原理)

题目链接

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=23842

题目大意

给出nn个数字a1,a2,a3...an

a1,a2,a3...an,对于所有存在的SS，求和为SS的三个数ai,aj,ak

ai,aj,ak的有序对(i,j,k)

(i,j,k) (i<j<k)(i<j<k)的个数。

思路

http://blog.csdn.net/qpswwww/article/details/44228381

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get new skill

在容斥的时候，如果三个多项式中两个多项式是绑定的（选两个重复的）

相当于两个多项式相加，也就是*2，同样三个多项式相同，乘以3

然后再进行卷积

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <math.h>
#include <memory.h>
#include <bits/stdc++.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const LL inf = 0x3f3f3f3f;
const LL MOD =100000000LL;
const int N = 350010;
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}


const double PI = acos(-1.0);
//复数结构体
struct Complex{
    double r,i;
    Complex(double _r = 0.0,double _i = 0.0){
        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);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂
 */
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]);
        //交换互为小标反转的元素，i<j保证交换一次
        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的
        k = len/2;
        while( j >= k){
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
/*
 * 做FFT
 * len必须为2^k形式，
 * on==1时是DFT，on==-1时是IDFT
 */
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;
}

Complex a[N],b[N];
int x1[N];
int x2[N];
int x3[N];
int main()
{
    int n,x;
    scanf("%d",&n);
    memset(x1,0,sizeof(x1));
    memset(x2,0,sizeof(x2));
    memset(x3,0,sizeof(x3));
    for(int i=0;i<n;i++)
    {
        scanf("%d",&x);
        x+=20000;
        x1[x]++;
        x2[x*2]++;
        x3[x*3]++;
    }
    int len=1<<17;
    for(int i=0;i<len;i++)
    {
        a[i]=Complex(x1[i],0);
        b[i]=Complex(x2[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]-Complex(3,0)*b[i]*a[i];
    }
    fft(a,len,-1);
    for(int i=0;i<len;i++)
    {
        x3[i]+=x3[i];
        x3[i]+=(int)(a[i].r+0.5);
        x3[i]/=6;
    }
    for(int i=0;i<len;i++)
    {
        if(x3[i]!=0)
        printf("%d : %d\n",i-60000,x3[i]);
    }
}