2015-02
Recursion and Mod
同餘
a\%M=a',\ b\%M=b'
a%M=a′, b%M=b′
(a+b)\%M=(a'+b')\%M
(a+b)%M=(a′+b′)%M
(a*b)\%M\ =(a'*b')\%M
(a∗b)%M =(a′∗b′)%M
題目敘述
Definition:
a_{1} = 1
a1=1
a_{n} = 3*a_{n-1} + 4
an=3∗an−1+4
解法一
就照著遞迴式做
a = 1
MOD = 10**9 + 9
for _ in range(N-1):
a = (3*a+4) % MOD時間複雜度:
O(N)=O(2^{63})=TLE
O(N)=O(263)=TLE
0.4
解法二
先觀察數列
a_{1}=1,a_{2}=7,a_{3}=25,a_{4}=79
a1=1,a2=7,a3=25,a4=79
似乎有甚麼規律
解法二
按照定義展開
a_{1}=1
a1=1
a_{2}=3*a_{1}+4
a2=3∗a1+4
a_{3}=3*a_{2}+4=3*(3*a_{1}+4)+1
a3=3∗a2+4=3∗(3∗a1+4)+1
解法二
a_{4}=3*a_{3}+4
a4=3∗a3+4
=3*(3*a_{2}+4)+4
=3∗(3∗a2+4)+4
=3*(3*(3*a_{1}+4)+4)+4
=3∗(3∗(3∗a1+4)+4)+4
=3^{3}*a_{1}+4*(3^{2}+3^{1}+3^{0})
=33∗a1+4∗(32+31+30)
=3^{3}+4*(3^3-1)/(3-1)
=33+4∗(33−1)/(3−1)
=3^{3}+2*3^{3}-2
=33+2∗33−2
=3^{4}-2
=34−2
解法二
再用數學歸納法仔細證明
a_{n}=3^{n}-2
an=3n−2
n=1 =>a_{1}=3^{1}-2=1
n=1=>a1=31−2=1
n=k+1=>a_{k+1}=3*a_{k}+4
n=k+1=>ak+1=3∗ak+4
=3*(3^{k}-2)+4
=3∗(3k−2)+4
=3^{k+1}-2
=3k+1−2
解法二
因此我們得知
a^{n}=3^n-2
an=3n−2
a = 1, MOD = 10**9 + 9
for _ in range(n):
a = (3*a) % MOD
a = a - 2時間複雜度:
O(N)=O(2^{63})=TLE
O(N)=O(263)=TLE
0.4
解法三
把遞迴式變成一般式
目標: 如何快速求得 ???
3^{n}
3n
快速冪
Fast Exponential Algorithm
快速冪
a^{n}={(a^{n/2})}^{2}
an=(an/2)2
={(a^{[n/2]})}^{2}*a
=(a[n/2])2∗a
if n is even
if n is odd
解法三
MOD = 10**9 + 9
def pow(a, n):
if n == 0:
return 1
half = pow(a, n//2)
if n&1 == 1:
return (((half**2)%MOD)*a)%MOD
else:
return (half**2)%MOD時間複雜度:
O(lgN)=O(lg2^{63})=O(63)
O(lgN)=O(lg263)=O(63)
AC
快速冪 迴圈版
ans = 1, A = a, MOD = 10**9 + 9
for i in range(32):
if (N>>i)&1:
ans = (ans*A) % MOD
A = (A*A) % MOD
GPE 5/18
By allenwhale
GPE 5/18
- 1,022
