Exemplo n.º 1
0
def check(x,y):
    if x > 2*y: return False
    d = pe.gcd(x,y)
    x/=d
    y/=d
    its = 0
    while y > 1:
        if x >= 2*y : return its%2
        x -= y
        its += 1
        x,y = max(x,y), min(x,y)
#        print x,y
    return its%2
Exemplo n.º 2
0
cap = 10**12
mod = 10**5

t = 1
while t <= cap:
  t5 = 1
  while t*t5 <= cap:
    h.append(t*t5)
    t5 *= 5
  t *= 2
    
o2 = prime_fact_ord(2,cap)
o5 = prime_fact_ord(5,cap)
exp2 = o2-o5 #Significant exponent of 2 in final product
h = sorted(h) #numbers divisible only by 2 and 5
print len(h)
res = 1 #init to one
for i in xrange(1,min(cap+1,mod)):
  if gcd(i,10) == 1:
    e = 0 #the exponent by i in the final product
    for m in h:
      if m*i > cap:
        break
      max_r = ((cap/m) - i) / mod #maximum r with m*(r*mod+i) <= cap
      e += max_r+1
    res = (res  * pow(i,e,mod)) % mod

res = (res * pow(2,exp2,mod)) % mod
print res
     
Exemplo n.º 3
0
primes,mask = pe.primes_and_mask(cap)

curr = np.ones(cap,dtype=np.int64)
for croak in croaks:
  prev = curr
  curr = np.ones(cap,dtype=np.int64)
  #Count croaks multiplicities. For each path on a prime, P gets croaked twice
  #For each on a non-prime, N gets croaked twice
  if croak == "P":
    for i in xrange(cap):
      if mask[i+1]:
        prev[i] *= 2
  else:
    for i in xrange(cap):
      if not mask[i+1]:
        prev[i] *= 2
  
  #compute transitions
  curr[1:cap-1] = prev[:cap-2] + prev[2:cap]
  curr[0] = prev[1]
  curr[cap-1] = prev[cap-2]
  #double count the contributions from the end squares
  curr[1] += prev[0]
  curr[cap-2] += prev[cap-1]   
num = np.sum(prev)   
print "Total admissible possibilities " + str(num)

d = pe.gcd(denom,num)
print "Reduced fraction " + str(num/d) + "/" + str(denom/d)