【SPOJ】Triple Sums【FFT】

2024-09-05 14:18

文章标签 triple fft spoj sums

本文主要是介绍【SPOJ】Triple Sums【FFT】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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：
$\qquad\qquad\qquad Y = \sum x^{a_i}$
那么 $ai+aj+ak=S$ 的个数即 $x^{a_i+a_j+a_k=S}$ 的个数，等价于 $Y^3中x^S$ 的系数。
然后我们考虑容斥：
$\qquad\qquad(\sum x)^3 = {\sum x^3+3\sum x^2y+6\sum xyz}$
$\qquad\qquad\sum x^2y = (\sum x^2)(\sum x)-\sum x^3$
$\qquad\qquad\sum xyz = \frac{(\sum x)^3 -3(\sum x^2)(\sum x)+2\sum x^3}{6}$

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 ;
}