HDU 2086 A1 = ?

/*
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)
      2[(A1+A2)+(C1+C2)] = A0+A2+A1+A3;
      A1+A2 = A0+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 <cstdio>
#include <iostream>
using namespace std;
const int maxn = 3003;
double a[maxn], c[maxn];
int main() {
    int n, i, j;
    while (scanf("%d", &n) == 1) {
        scanf("%lf%lf", &a[0], &a[n + 1]);
        for (i = 1; i <= n; i++) 
            scanf("%lf", &c[i]);
        a[1] = n * a[0] + a[n + 1];
        double sum = 0;
        for (i = n, j = 1; i >= 1 && j <= n; j++, i--) 
            sum += i * c[j];
        a[1] = (a[1] - 2 * sum) / (n + 1);
        printf("%.2lf\n", a[1]);
    }
    return 0;
}

//递推题：        
//a[1]=1/(n+1)*(a[n+1]+n*a[0]-2*c[n]-4*c[n-1]-8*c[n-2]...-pow(2,n)*c[1]);

#include<stdio.h>

int main()
{
    int n,i,k;
    float a[3005],c[3005];
    float sum;
    while(~scanf("%d",&n))
    {
        scanf("%f%f",&a[0],&a[n+1]);
        for(i=1;i<=n;++i)
            scanf("%f",&c[i]);
        k=2;
        sum=0;
        for(i=n;i>0;--i)
        {
            c[i]*=k;
            k+=2;
            sum+=c[i];
        }
        printf("%.2f\n",1.0/(n+1)*(a[n+1]+n*a[0]-sum));
    }
    return 0;
}

//n = 1时，有 a1 = a0/2 + a2/2 - c1
//n = 2时，有 3/2*a1 = a0 + a3/2 - 2*c2 -  c1
//n = 3时，有 2*a1 = 3*a0/2 + a4/2 - 3*c3 - 2*c2 -  c1
//所以得出
//n = k 时，有 (n+1)/2 * a1 = (n/2)*a0 + a(n+1)/2 - n*cn - …-2*c2 - c1
#include <stdio.h>
int main(){
    int n,i;
    double a0,an,c[3005];
    while(~scanf("%d",&n)){
        scanf("%lf%lf",&a0,&an);
        for(i = 1;i<=n;i++)
        scanf("%lf",&c[i]);
        double ans,t = 0;
        for(i = 1;i<=n;i++)
            t = t+(n-i+1)*c[i];
        ans = a0*(0.5*n)+an/2.0-t;
        ans/=(0.5*(n+1));
        printf("%.2lf\n",ans);
    }

    return 0;
}