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A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4682    Accepted Submission(s): 2990

1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00

27.50
15.00

分析：

公式推导

A₅ = (A₄ + A₆) / 2 - C₅

A₄ = (A₃ + A₅) / 2 - C₄ = A₃ / 2 + A₄ / 4 + A₆ / 4 - C₅ / 2 - C₄
=>A₄ = 2A₃/3 + A₆/3 - 2C₅/3 - 4C₄/3

A₃ = (A₂ + A₄) / 2 - C₃ = A₂ / 2 + A₃ / 3 + A₆ / 6 - C₅ / 3 - 2C₄ / 3 - C₃
=>A₃ = 3A₂/4 + A₆/4 - C₅/2 - C₄ - 3C₃/2

A₂ = (A₁ + A₃) / 2 - C₂ = A₁ / 2 + 3A₂ / 8 + A₆ / 8 - C₅ / 4 - C₄ / 2 - 3C₃ / 4 - C₂
=>A₂ = 4A₁/5 + A₆/5 - 2C₅/5 - 4C₄/5 - 6C₃/5 - 8C₂/5

A₁ = (A₀ + A₂) / 2 - C₁ = A₀ / 2 + 2A₁ / 5 + A₆ / 10 - C₅ / 5 - 2C₄ / 5 - 3C₃ / 5 - 4C₂ / 5 - C₁
=>A₁ = 5A₀/6 + A₆/6 - C₅/3 - 2C₄/3 - C₃ - 4C₂/3 - 5C₁/3

算到这里，我想你已经总结出公式了:
A1 = (n * A₀ + A_n+1 - 2 * C_n - 4 * C_n-1 - ... - 2 * i * C_n-i+1 - 2 * n * C₁) / (n + 1)

<span style="font-size:18px;">#include<stdio.h>
int main ()
{
  int t,i;
  double a,b,c[10000],d;
  while (~scanf("%d",&t))
  {
    scanf("%lf%lf",&a,&b);
	for(i=1;i<=t;i++)
		scanf("%lf",&c[i]);
    d=t*a+b;
	for(i=1;i<=t;i++)
       d-=2*i*c[t-i+1];
	printf("%.2lf\n",d/(t+1));
  }
  return 0;
}</span>

hdu 2086 A1 = ?,布布扣,bubuko.com

hdu 2086 A1 = ?

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原文地址：http://blog.csdn.net/fyxz1314/article/details/29622295