PS:推算。。。数组如果开得不够大也会超时。。。

代码:

#include "stdio.h"double cal(int t,double a[]);int main(){    double a0,an1,c[1100],sum,k;    int i,n,b;    while(~scanf("%d",&n)){        scanf("%lf%lf",&a0,&an1);        for(i=1;i<=n;i++)            scanf("%lf",&c[i]);        sum=(double)n/(double)(n+1)*a0+1.0/(n+1)*an1;        k=(double)n*2.0;        for(i=1;i<=n;i++){            sum-=k/(n+1)*c[i];            k-=2.0;        }        printf("%.2lf\n",sum);    }    return 0;}