def pe202():
"""
| \| \| \| \| \| \| Expand triangle mirrors infinitely, then every
+--+--+--+--+--+--+- possible path is a C-C line from bottomleft to
C|\B|\A|\C|\B|\A|\C|\ somewhere upright and doesn't cross any vertex.
| \| \| \| \| \| \|
+--+--+--+--+--+--+- From the table can easity see that C-C lines are
B|\A|\C|\B|\A|\C|\B|\ symmetric and lie on y = x + 3k. Without loss of
| \| \| \| \| \| \| generality, we just need to count all (a, b) with
+--+--+--+--+--+--+- a > b.
A|\C|\B|\A|\C|\B|\A|\
| \| \| \| \| \| \| Also, from (0, 0) to (a, b), the amount of cross
+--+--+--+--+--+--+- points with horizontal, vertical and diagonal
C B A C B A C lines is (a-1) + (b-1) + (a+b-1) = 2a + 2b - 3.
Therefore to bounce N times, we need to find all (a, b) that:
(1) a + b = (N + 3) / 2
(2) gcd(a, b) = 1
(3) a - b = 3k
Recording to reference articles PE202, the number of solutions of (1), (2)
and (3) is:
sum m(d) * f(n/d) for all d satisfying d|n
where m(x) if mobius function, f(x) = floor(x/2) - floor(x/3)
"""
def f(x):
return x / 2 - x / 3
N = 12017639147
N = (N + 3) / 2 # 6008819575
# 6008819575 = 5*5 * 11 * 17 * 23 * 29 * 41 * 47
pfactors = (5, 11, 17, 23, 29, 41, 47)
count = f(N)
for i in range(1, len(pfactors)+1):
for pcomb in pec.combinations(pfactors, i):
d = pef.cprod(pcomb)
if i % 2:
count -= f(N / d)
else:
count += f(N / d)
# answer: 1209002624
return count * 2
def pe209():
"""
Take input as integer in 0~63. Then it's equivalent to how many binary
functions F(x) with 0 <= x <= 63 subjected to F(a) * F(b) = 0 for some
a, b which b is determined by a.
There are some cycles in the mapping. So we just need to find valid
ways of each cycle, which becomes a easy combinatoric problem.
"""
def valid_permutations(n):
"""
Count valid permutations of chain with length n. Each element of chain
can only be 0 or 1. No consecutive 1s. Head and tail is regarded as
consecutive too.
Therefore it would be 0~floor(n/2) possible 1s.
For a chain of A 0s and B 1s, there's C(A-1, B-1) chains with head of 1
and C(A, B) chains with head of 0.
"""
x = 1
for i in range(1, n/2+1):
x += pec.C(n-i-1, i-1) + pec.C(n-i, i)
return x
pairs = {}
for n in range(64):
s = bin(n)[2:]
s = '0' * (6-len(s)) + s
a, b, c = map(int, s[:3])
s = s[1:] + str(a ^ (b & c))
pairs[n] = int(s, 2)
# There are 6 cycles in pairs, started with 0, 1, 3, 5, 9, 21
starts, lchains = (0, 1, 3, 5, 9, 21), []
for i in starts:
z = set([i])
n = pairs[i]
while n > i:
z.add(n)
n = pairs[n]
lchains.append(len(z))
# answer: 15964587728784
return pef.cprod(map(valid_permutations, lchains))
def F(plist):
return pef.cprod([2*n+1 for n in plist])
