HDU 2086 A1 =?

A1 =? Time limit:5000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) Total submission (s): 6492 Accepted Submission (s): 4038 Problem Description There are the following equations: ai = (ai-1 + ai+1)/2-ci (i = 1, 2, 3, .... n). If given a0, an+1, and C1, C2, ..... Cn. Please programmatically calculate a1 =? Input The input includes multiple test instances. For each instance, the first is a positive integer n, (n <= 3000); Then there are 2 numbers a0, an+1. The next n rows have a number Ci(i = 1, ...) for each row, and the input ends with a file terminator. Output For each test instance, a row is used to output the calculated A1 (retain 2 decimal places). Sample Input 1 50.00 25.00 10.00 2 50.00 25.00 10.00 20.00 Sample Output 27.50 15.00 Source 2006/1/15 ACM Program Design Final Exam

1 /* 2 ai= (ai-1+ai+1)/2-ci,3 a1= (A0 + A2)/2-c1;4 a2= (A1 + A3)/2-c2, ...5 = = A1+a2 = (A0+A2+A1+A3)/2-(C1+C2)6 2[(A1+A2) + (C1+C2)] = A0+A2+A1+A3;7 a1+a2 = a0+a3-2 (C1+C2);8 = = A1+A2 = A0+a3-2 (C1+C2)9 In the same vein, you get:Ten a1+a1 = a0+a2-2 (C1) One a1+a2 = a0+a3-2 (C1+C2) A a1+a3 = a0+a4-2 (C1+C2+C3) - a1+a4 = a0+a5-2 (C1+C2+C3+C4) -       ...  the A1+an = a0+an+1-2 (C1+C2+...+CN) - sum-----------------------------------------------------left and right - (n+1) a1+ (A2+a3+...+an) = nA0 + (A2+a3+...+an) + an+1-2 (nc1+ (n-1) C2+...+2CN-1+CN) -    + = = (n+1) A1 = nA0 + an+1-2 (nc1+ (n-1) C2+...+2CN-1+CN) -    + = = A1 = [nA0 + an+1-2 (nc1+ (n-1) C2+...+2CN-1+CN)]/(n+1) A */   at //!!! With CIN and cout will time out -#include <iostream> -#include <cstdio> - using namespacestd; - Doublea[3010],c[3010]; - intMain () in { -     intN; to      while(~SCANF ("%d",&N)) { +         Doublesum=0; -scanf"%LF%LF", &a[0],&a[n+1]); the          for(inti=n;i>=1; i--){ *scanf"%LF",&c[i]); $sum+=i*C[i];Panax Notoginseng         } -a[1]= (a[0]*n+a[n+1]-2*sum)/(n+1); theprintf"%.2f\n", a[1]); +     } A     return 0; the}

HDU 2086 A1 =?