学步园

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

解题报告

拿笔模拟下就出来了。。。

2a2=a1+a3+2c2;

2a3=a2+a4+2c3;
2a4=a3+a5+2c4;
2a5=a4+a6+2c5;
累加就出来了。。。
#include<stdio.h>
int main()
{
    int i,n,t,j;
    double a0,a1,at,c[5000],ct;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        scanf("%lf%lf",&a0,&at);
        for(i=1;i<=n;i++)
        scanf("%lf",&c[i]);
        ct=0;
        for(i=1,j=n;i<=n;i++,j--)
        ct+=j*c[i];
        a1=(at+n*a0-2*ct)/(n+1);
        printf("%.2lf\n",a1);
        if(t)
        printf("\n");
    }
    return 0;
}