Simple calculations
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
题意：由公式ai = (ai-1 + ai+1)/2 - ci ；
给出 a0 an+1 和 c1，c2，c3......ci,求出a1
方法：
2an = an-1 + an+1 - 2cn ；等价于
2an-1 = an-2 + an - 2cn-1 ；等价于
. . . .
. . . .
. . . .
2a3 =
a2
+
a4
- 2c3 ；等价于
2a2 =
a1
+
a3
- 2c2 ；等价于
2a1 = a0 + a2 - 2c1 ；
相加得出：
an+1 - a1 = an- a0 + 2(c1+c2+c3+....+cn)
因为
还不知道，所以要转化：
an+1 - an = a1- a0 + 2(c1+c2+c3+..............+cn)
an - an-1 = a1- a0 + 2(c1+c2+c3+..........+cn-1)
an-1 - an-2 = a1- a0 + 2(c1+c2+c3+....+cn-2)
. . . . . . .
. . . . . . .
. . .. . . . .
a3 - a2 = a2- a1 + 2(c1 +
a2 - a1 = a1- a0 + 2c1
相加化简得：
an+1 - a1= a1- a0 +2( nc1 + (n-1)*c2 + (n-2)c3 +.....+2cn-1+cn )
化简得：
a1=(an+1 +na0-2( nc1 + (n-1)*c2 + (n-2)c3 +.....+2cn-1+cn ))/(n+1)
1 #include <cstdlib>
2 #include <iostream>
3
4 using namespace std;
5 float c[3001],a[3002];
6 int main(int argc, char *argv[])
7 {
8 int n,i,j; float cc=0;
9 cin>>n;
10 cin>>a[0]>>a[n+1];
11 for(i=1;i<=n;i++)
12 {
13 cin>>c[i];
14 cc+=(n-i+1)*c[i];
15 }
16 a[1]=(a[n+1]+n*a[0]-2*cc)/(n+1);
17 printf("%.2f",a[1]);
18 system("PAUSE");
19 return EXIT_SUCCESS;
20 }
21