Simple calculations(数学都不会了)

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 - ci for 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 should contain a1 in the same format as a0 and an+1.

Sample Input
1
50.50
25.50
10.15

Sample Output
27.85

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csdn renyuzhuo0

由原式得：

2a[i]=a[i-1]+a[i+1]-c[i]

所以

2a[i]  =a[i-1]+a[i+1]-c[i]

2a[i-1]=a[i-2]+a[i]-c[i-1]

2a[i-2]=a[i-3]+a[i-1]-c[i-2]

：

2a[2]=a[1]+a[3]-c[2]

2a[1]=a[0]+a[2]-c[1]

左右两侧分别相加得：

a[i]+a[1]=a[0]+a[i+1]-(c[1]+c[2]+...+c[i-2]+c[i-1]+c[i])

即:c[1]+c[2]+...+c[i-2]+c[i-1]+c[i]=a[0]+a[i+1]-a[i]-a[1]

所以

a[0]+a[i+1]-a[i]-a[1]=c[1]+c[2]+...+c[i-2]+c[i-1]+c[i]

a[0]+a[i]-a[i-1]-a[1]=c[1]+c[2]+...+c[i-2]+c[i-1]

a[0]+a[i-1]-a[i-2]-a[1]=c[1]+c[2]+c+c[i-2]

：

a[0]+a[3]-a[2]-a[1]=c[1]+c[2]

a[0]+a[2]-a[1]-a[1]=c[1]

左边右边分别相加：

化简得：

n*a[0]+a[i+1]-(n+1)*a[1]=n*c[1]+(n-1)*c[2]+...c[n]

a[1]=(n*a[0]+a[i+1]-(n*c[1]+(n-1)*c[2]+...c[n]))/(n+1)

此时可看出，除了a[1],其他都已知，可编代码：

#include<stdio.h>#include<string.h>int main(){    int n, i;    double first, final, sum, temp, ans;    while(scanf("%d",&n)!=EOF)    {        scanf("%lf", &first);        scanf("%lf", &final);        for(i=1,sum=0; i<=n; ++i)        {            scanf("%lf", &temp);            sum += 2.0*(n+1-i)*temp; //c数组求和        }        ans = n*first+final-sum;        printf("%.2lf\n", ans/(n+1));    }    return 0;}