UVA10014 - Simple calculations

公式：a1 = n * a0 / (n + 1) + an+1 / (n + 1) + ∑∑ci

题目：

Simple calculations

 Simple calculations 

The Problem

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.

The Input

The first line is the number of test cases, followed by a blank line.

For each test case, the first line of an input file 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.

Each test case will be separated by a single line.

The Output

For each test case, the output file should contain a₁ in the same format as a₀ and a_n+1.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

1

1
50.50
25.50
10.15

Sample Output

27.85

解答：

#include<stdio.h>
int main()
{
    int t;
    double a0,an1,c;
    scanf("%d",&t);
    while(t--)
    {
        int i,j;
        double n,c,s=0;
        scanf("%lf%lf%lf",&n,&a0,&an1);
        for(i=0;i<n;i++)
        {
            scanf("%lf",&c);
            s+=(n-i)*c;
        }
        s*=2;
        printf("%.2lf\n",n*a0/(n+1)+an1/(n+1)-s/(n+1));
        if(t)
            printf("\n");
    }
    return 0;
}