Jzzhu and Apples (Codeforces 449C)
Tags: Others, number theory

Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to an apple store. Jzzhu will pack his apples into groups and then sell them. Each group must contain two apples, and the greatest common divisor of numbers of the apples in each group must be greater than 1. Of course, each apple can be part of at most one group.Jzzhu wonders how to get the maximum possible number of groups. Can you help him?

Input: A single integer n (1 ≤ n ≤ 105), the number of the apples.

Output: The first line must contain a single integer m, representing the maximum number of groups he can get. Each of the next m lines must contain two integers — the numbers of apples in the current group.If there are several optimal answers you can print any of them.


Sample Inputs:
6

9

2

Sample Outputs:
2
6 3
2 4

3
9 3
2 4
6 8

0

Solution (Python):
apples = int(input())
if apples <= 3:
    print(0)
else:
    halfpr = int(apples / 2)

    def primes(n):
        isPrime = [True for i in range(n + 1)]
        isPrime[0] = isPrime[1] = False
        idx = 2
        while idx * idx <= n:
            if isPrime[idx]:
                for i in range(idx * 2, n, idx):
                    isPrime[i] = False
            idx += 1
        return isPrime
    primeslist = primes(halfpr)
    totallist = [False] * (apples + 1)
    applepairs = []
    for prime in range(len(primeslist) - 1, 1, -1):
        if primeslist[prime]:
            numprimes = int(apples / prime)
            primesx = [int(i * prime) for i in range(1, numprimes + 1) if not totallist[i * prime]]
            if len(primesx) % 2 == 1:
                primesx.remove(2 * prime)
            for pr in primesx:
                applepairs.append(pr)
                totallist[pr] = True
    print(int(len(applepairs) / 2))
    for t in range(int(len(applepairs) / 2)):
        print(applepairs[2 * t], applepairs[2 * t + 1])

Solution (Python):
""""""

def read(mode=2):
    inputs = input().strip()
    if mode == 0:
        return inputs
    if mode == 1:
        return inputs.split()
    if mode == 2:
        return [int(x) for x in inputs.split()]

def write(s='\n'):
    if isinstance(s, list):
        s = ' '.join(map(str, s))
    s = str(s)
    print(s, end='')
from itertools import compress
import itertools

def croft():
    """"""
    for p in (2, 3, 5):
        yield p
    roots = {9: 3, 25: 5}
    primeroots = frozenset((1, 7, 11, 13, 17, 19, 23, 29))
    selectors = (1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0)
    for q in compress(itertools.islice(itertools.count(7), 0, None, 2), itertools.cycle(selectors)):
        if q in roots:
            p = roots[q]
            del roots[q]
            x = q + 2 * p
            while x in roots or x % 30 not in primeroots:
                x += 2 * p
            roots[x] = p
        else:
            roots[q * q] = q
            yield q
(n,) = read()
cr = croft()
primes = []
for i in cr:
    if i < n:
        primes.append(i)
    else:
        break
primes.reverse()
used = [0] * (n + 1)
res = []
for p in primes:
    k = n // p
    tmp = []
    while k:
        if not used[k * p]:
            tmp.append(k * p)
            used[k * p] = 1
        if len(tmp) == 2:
            res.append(tmp)
            tmp = []
        k -= 1
    if tmp == [p] and p > 2 and (p * 2 <= n) and len(res) and (res[-1][1] == p * 2):
        res[-1][1] = p
        used[p * 2] = 0
        used[p] = 1
print(len(res))
for i in res:
    print(' '.join(map(str, i)))

Solution (Python):
from math import *

def prm(x):
    s = int(sqrt(x + 0.5))
    for i in range(2, s + 1):
        if not x % i:
            return False
    return True

def ap(a):
    while len(a) >= 2:
        print(a[-1], a[-2])
        a.pop()
        a.pop()
(n, cnt, ans, vis, ansl, ansr) = (int(input()), 0, 0, [False] * 100100, [], [])
prime = filter(prm, range(3, n // 2 + 1))
for x in prime:
    (tcnt, nlst) = (0, [])
    for i in range(1, n // x + 1):
        if not vis[i * x]:
            vis[i * x] = True
            nlst.append(i)
            tcnt += 1
    ans += tcnt >> 1
    cnt += tcnt & 1
    if tcnt & 1:
        nlst = nlst[0:1] + nlst[2:]
        ansr.append(x << 1)
    ansl += list(map(lambda k: x * k, nlst))
for i in range(1, n + 1):
    if not vis[i] and (not i & 1):
        ansr.append(i)
        cnt += 1
print(ans + (cnt >> 1))
ap(ansl)
ap(ansr)