hdu2086A1 = ?

http://acm.hdu.edu.cn/showproblem.php?pid=2086

推出公式为（  n + 1 )  *  a1 + 2 * ( c1 + c1+c2+cq+c2+c3+......+cn) = n * a0 + an 

#include "stdio.h"
#include "string.h"
#include "stdlib.h"
#include "math.h"
#include "algorithm"
#include "iostream"

using namespace std;

int main()
{
	int n , i , j ; 
	double a0 , an ,c[ 3005 ] ,temp ,sum1 ,sum;
	while( ~scanf( "%d" , &n ) )
	{
		scanf( "%lf" ,&a0 ) ;
		scanf( "%lf", &an ) ;
		sum1 = sum = 0 ;
		for(  i = 1 ; i <= n ; i++ )
		{
			scanf( "%lf" , &temp) ;
			sum1 += temp ;
			sum += sum1 * 2.0 ;
		/*	scanf( "%lf" ,&c[ i ] ) ;
		double temp = 0.0 ;
		for(  i = 1 , j = n ; i <= n ; i++ ,j-- )
		{
			temp +=  i * c[ j ] ;
		}
		temp = 2.0 * temp ;*/
		}
		printf( "%.2lf\n" , 1.0 * ( ( 1.0 * n * a0 + an ) - sum ) /( 1.0 *( n + 1 ) ) ) ;
	}
	return 0;
}