首先变形为:
a[i + 1] = 2 * a[i] - a[i - 1] + 2 * c[i]

比如
a[2] = 2 * a[1] - a[0] + 2 * c[1]
发现它是a[1]的线性组合，猜想a[3]也是a[1]的线性组合

那么设
a[i] = p[i] * a[1] + q[i]
则
a[i + 1] = 2 * a[i] - a[i - 1] + 2 * c[i]
         = 2 * (p[i] * a[1] + q[i]) - (p[i - 1] * a[1] + q[i - 1]) + 2 * c[i]
         = (2 * p[i] - p[i - 1]) * a[1] + (2 * q[i] - q[i - 1] + 2 * c[i])
即
p[i + 1] = 2 * p[i] - p[i - 1]
q[i + 1] = 2 * q[i] - q[i - 1] + 2 * c[i]
由此递推得到a[n + 1]关于a[1]的线性表达式，即可求解

• 顶这个，比较有技巧，比硬推强得多。 (0) oipn4e2 2009-07-31 22:40:02
• UPUPUP (513) moretimes 2012-10-17 09:27:43
  • UP&ORZ moretimes (0) Tune 2013-03-10 16:43:07
• Re:我的做法 (5) HermioneS 2014-06-09 14:44:20