递推

斐波那契数列：从前往后推

f(i)=f(i-1)+f(i-2)

k = int(input())
f = [0] * (k+2)
f[1] = 1
f[2] = 1
for i in range(3, k+1):
    f[i] = f[i-1] + f[i-2]
print(f[k])

next数组，寻找下一个非1的元素：从后往前推

f(i) = i+1, if a(i+1)!=1.

f(i) = f(i+1), if a(i+1)==1

练习题

http://oj.oldmoon.cn/p/NOIPJ2015A

差分思想

常见调试技巧