PKU-2601-Simple calculations

Description

There is a sequence of n+2 elements a₀, a₁, ..., a_n+1 (n <= 3000, -1000 <= a_i <=1000). It is known that a_i = (a_i-1 + a_i+1)/2 - c_i for each i=1, 2, ..., n.
You are given a₀, a_n+1, c₁, ... , c_n. 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 a_n+1 each having two digits after decimal point, and the next n lines contain numbers c_i (also with two digits after decimal point), one number per line.

Output

The output file should contain a₁ in the same format as a₀ and a_n+1.

Sample Input

1
50.50
25.50
10.15

Sample Output

27.85

推公式呀推公式~~~

a[1] + c[1] = 1/2 (a[0] + a[2])
a[2] + c[2] = 1/2 (a[1] + a[3])
...
a[n] + c[n] = 1/2 (a[n-1] + a[n+1])
then..
a[1..n] + c[1..n] = 1/2 (a[0..n-1] + a[2..n+1])
 
a[1] + a[n]  = a[0] + a[n+1] -2c[1..n]
a[1] + a[n-1]= a[0] + a[n]   -2c[1...n-1]
...
a[1] + a[1]  = a[0] + a[2]   -2c[1..1]
then..
na[1]+a[1..n]=na[0]+a[2..n+1]-2c[....]      c[....]={nc[1]+(n-1)c[2]+...+c[n]}
finally..
(n+1)a[1]=na[0]+a[n+1]-2c[...]
#include <stdio.h>

int main()
{
    int n,i;
    double a,b,temp;
    double sum=0.0;
//    freopen("d:\\input.txt","r",stdin);
    scanf("%d",&n);
    scanf("%lf%lf",&a,&b);
    for (i=0;i<n;i++)
    {
        scanf("%lf",&temp);
        sum+=(n-i)*temp;
    }
    sum=-sum*2;
    sum+=n*a+b;
    sum/=n+1;
    printf("%.2lf\n",sum);
    return 0;
}