【SPOJ】Triple Sums【FFT】

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传送门：【SPOJ】Triple Sums

题目分析：

首先我们不考虑i<j<k这个条件，构造多项式：
Y=∑xai
那么ai+aj+ak=S的个数即xai+aj+ak=S的个数，等价于Y3中xS的系数。
然后我们考虑容斥：
(∑x)3=∑x3+3∑x2y+6∑xyz
∑x2y=(∑x2)(∑x)−∑x3
∑xyz=(∑x)3−3(∑x2)(∑x)+2∑x36

然后FFT处理下，这道题就做完啦。

my code：

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 200005 ;const double pi = acos ( -1.0 ) ;struct Complex {    double r , i ;    Complex () {}    Complex ( double r , double i ) : r ( r ) , i ( i ) {}    Complex operator + ( const Complex& t ) const {        return Complex ( r + t.r , i + t.i ) ;    }    Complex operator - ( const Complex& t ) const {        return Complex ( r - t.r , i - t.i ) ;    }    Complex operator * ( const Complex& t ) const {        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;    }} ;void FFT ( Complex y[] , int n , int rev ) {    for ( int i = 1 , j , k , t ; i < n ; ++ i ) {        for ( j = 0 , k = n >> 1 , t = i ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;        if ( i < j ) swap ( y[i] , y[j] ) ;    }    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {        Complex wn ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {            for ( int i = k ; i < n ; i += s ) {                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;                y[i] = y[i] + t ;            }        }    }    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;}int n ;int num[MAXN] ;Complex x1[MAXN] ;Complex x2[MAXN] ;Complex x3[MAXN] ;void solve () {    int x , n1 = 1 ;    clr ( num , 0 ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        scanf ( "%d" , &x ) ;        num[x + 20000] ++ ;    }    while ( n1 <= 80000 ) n1 <<= 1 ;    for ( int i = 0 ; i < n1 ; ++ i ) x1[i] = Complex ( num[i] , 0 ) ;    for ( int i = 0 ; i < n1 ; ++ i ) {        if ( i % 2 == 0 ) x2[i] = Complex ( num[i / 2] , 0 ) ;        else x2[i] = Complex ( 0 , 0 ) ;    }    FFT ( x1 , n1 , 1 ) ;    FFT ( x2 , n1 , 1 ) ;    for ( int i = 0 ; i < n1 ; ++ i ) x3[i] = x1[i] * x1[i] * x1[i] ;    for ( int i = 0 ; i < n1 ; ++ i ) x2[i] = x2[i] * x1[i] ;    FFT ( x3 , n1 , -1 ) ;    FFT ( x2 , n1 , -1 ) ;    for ( int i = 0 ; i < n1 ; ++ i ) {        LL ans = ( LL ) ( x3[i].r + 0.5 ) - 3 * ( LL ) ( x2[i].r + 0.5 ) ;        if ( i % 3 == 0 ) ans += 2LL * num[i / 3] ;        ans /= 6 ;        if ( ans ) printf ( "%d : %lld\n" , i - 60000 , ans ) ;    }}int main () {    while ( ~scanf ( "%d" , &n ) ) solve () ;    return 0 ;}

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