Poj 2601 Simple calculations

Poj 2601 Simple calculations

Simple calculations

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:6559		Accepted:3291

Description

There is a sequence of n + 2 elements a0, a1,..., an + 1 (n <= 3000,-1000 <= ai <= 1000). It is known that Ai = (ai-1 + ai + 1)/2-ciFor each I = 1, 2,..., n.
You are given a0, an + 1, c1,..., cn. Write a program which calculates a1.

Input

The first line of an input contains an integer n. the next two lines consist of numbers a0 and an + 1 each having two digits after decimal point, and the next n lines contain numbers ci (also with two digits after decimal point ), one number per line.

Output

The output file shoshould contain a1 in the same format as a0 and an + 1.

Sample Input

150.5025.5010.15

Sample Output

27.85

Solution:

General process: a [0] + a [2]-2a [1]-2c [1] = 0a [1] + a [3]-2a [2]-2c [2] = 0 ...... A [n-1] + a [n + 1]-2a [n]-2c [n] = 0 accumulate: a [0] + a [n + 1]-a [1]-a [n]-2c [1]-2c [2]-... -2c [n] = 0 based on a [n-1] + a [n + 1]-2a [n]-2c [n] = 0 => a [n + 1]- 2c [n]-a [n] = a [n] + 2c [n]-a [n-1] simplification: a [0] + a [n]-a [1]-a [n-1]-2c [1]-2c [2]-... -Similarly, 2c [n-1] = 0: a [0] + a [n-1]-a [1]-a [N-2]-2c [1]-2c [2]-... -2c [N-2] = 0 ...... A [0] + a [2]-a [1]-a [1]-2c [1] = 0 add all the above types to obtain n * a [0] + a [n + 1]-(n + 1) * a [1]-2 * n * c [1]-2 * (n-1) * c [2]-... -2 * c [n] = 0: a [1] = (n * a [0] + a [n + 1]-2 * n * c [1]-2 * (n-1) * c [2]-... -2 * c [n])/(n + 1)

#include 
 
  #include 
  
   using namespace std;#define MAX 3005int main(){int n;double a0,an;double c[MAX];while (cin>>n){cin>>a0>>an;double ans=0;for (int i=0;i
   
    >c[i];ans+=2*(n-i)*c[i];}ans=(n*a0+an-ans)/(n+1);cout<