SOJ-2857（数学递推公式）

/****************************************************************************************************** ** Copyright (C) 2011.07.01-2013.07.01 ** Author: famousDT <[e-mail]> ** Edit date: 2011-07-17******************************************************************************************************/#include <stdio.h>#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10#include <vector>#include <queue>#include <map>#include <time.h>#include <set>#include <stack>#include <string>#include <iostream>#include <assert.h>#include <string.h>//memcpy(to,from,count#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll#include <algorithm>using namespace std;//typedef long long ll;#define PI acos(-1)#define MAX(a, b) ((a) > (b) ? (a) : (b))#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))#define FABS(a) ((a) >= 0 ? (a) : (-(a)))#define MAX_INT 0xfffffff/***********************************解题报告******************************************* ** a1 = (a0 + a2) / 2 -c1    ->    2 * a1 = a0 + a2 - 2 * c1 ** a2 = (a1 + a3) / 2 -c2    ->    2 * a2 = a1 + a3 - 2 * c2 ** a3 = (a2 + a4) / 2 -c3    ->    2 * a3 = a2 + a4 - 2 * c3 ** ……    ……    …… ** an = (an-1 + an+1) / 2 -cn    ->    2 * an = an-1 + an+1 - 2 * cn ** 保留第一行，并从第一个式子往下加 ** 2 * a1 = a0 + a2 - 2 * c1 ** a1 + a2 = a0 + a3 - 2 * (c1 + c2) ** a1 + a3 = a0 + a4 - 2 * (c1 + c2 + c3) ** ……    ……    …… ** a1 + an = a0 + an+1 - 2 * (c1 + c2 + c3 + c4 + … + cn) ** n个式子相加，得 ** (n + 1) * a1 = n * a0 + an+1 - 2 * (n * c1 + (n - 1) * c2 + (n - 2) * c3 + … + cn)****************************************************************************************/int main(){    int n;    int i;    double a, b;    while (scanf("%d", &n) == 1) {        scanf("%lf%lf", &a, &b);        double ans, c;        ans = n * a + b;        for (i = 1; i <= n; ++i) {            scanf("%lf", &c);            ans -= 2 * (n - i + 1) * c;        }        printf("%.2lf\n", ans / (n + 1));    }    return 0;}