hdu2086

A1 = ?
Time Limit: 1000msMemory Limit: 32768KB This problem will be judged on HDU. Original ID: 2086
64-bit integer IO format: %I64d Java class name: Main
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有如下方程：Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, …. n).
若给出A0, An+1, 和 C1, C2, …..Cn.
请编程计算A1 = ?
Input
输入包括多个测试实例。
对于每个实例，首先是一个正整数n,(n <= 3000); 然后是2个数a0, an+1.接下来的n行每行有一个数ci(i = 1, ….n);输入以文件结束符结束。
Output
对于每个测试实例，用一行输出所求得的a1(保留2位小数).
Sample Input
1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00
Sample Output
27.50
15.00
Source
Basic Problem

因为：Ai=(Ai-1+Ai+1)/2 - Ci,
A1=(A0 +A2 )/2 - C1;
A2=(A1 + A3)/2 - C2 , …
=> A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)
=> A1+A2 = A0+A3 - 2(C1+C2)
同理可得：
A1+A1 = A0+A2 - 2(C1)
A1+A2 = A0+A3 - 2(C1+C2)
A1+A3 = A0+A4 - 2(C1+C2+C3)
A1+A4 = A0+A5 - 2(C1+C2+C3+C4)
…
A1+An = A0+An+1 - 2(C1+C2+…+Cn)
—————————————————– 左右求和
(n+1)A1+(A2+A3+…+An) = nA0 +(A2+A3+…+An) + An+1 - 2(nC1+(n-1)C2+…+2Cn-1+Cn)

=> (n+1)A1 = nA0 + An+1 - 2(nC1+(n-1)C2+…+2Cn-1+Cn)

=> A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+…+2Cn-1+Cn)]/(n+1)
*/

#include<stdio.h>int main(){    int t,i,n;    double a,b,c,a1,sum1,sum2;    while(scanf("%d",&n)!=EOF)    {        t=n;        scanf("%lf%lf",&a,&b);        sum1=t*a+b;        sum2=0;        for(i=1;i<=n;i++)        {            scanf("%lf",&c);            sum2=sum2+t*c;            t--;        }        a1=(sum1-2*sum2)/(n+1);        printf("%.2f\n",a1);    }    return 0;}