Competitive Programmer (Codeforces 1266A)
Tags: Math, chinese remainder theorem

Bob is a competitive programmer. He wants to become red, and for that he needs a strict training regime. He went to the annual meeting of grandmasters and asked $$$n$$$ of them how much effort they needed to reach red."Oh, I just spent $$$x_i$$$ hours solving problems", said the $$$i$$$-th of them. Bob wants to train his math skills, so for each answer he wrote down the number of minutes ($$$60 \cdot x_i$$$), thanked the grandmasters and went home. Bob could write numbers with leading zeroes — for example, if some grandmaster answered that he had spent $$$2$$$ hours, Bob could write $$$000120$$$ instead of $$$120$$$.Alice wanted to tease Bob and so she took the numbers Bob wrote down, and for each of them she did one of the following independently:   rearranged its digits, or  wrote a random number. This way, Alice generated $$$n$$$ numbers, denoted $$$y_1$$$, ..., $$$y_n$$$.For each of the numbers, help Bob determine whether $$$y_i$$$ can be a permutation of a number divisible by $$$60$$$ (possibly with leading zeroes).

Input: The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 418$$$) — the number of grandmasters Bob asked.Then $$$n$$$ lines follow, the $$$i$$$-th of which contains a single integer $$$y_i$$$ — the number that Alice wrote down.Each of these numbers has between $$$2$$$ and $$$100$$$ digits '0' through '9'. They can contain leading zeroes.

Output: Output $$$n$$$ lines.For each $$$i$$$, output the following. If it is possible to rearrange the digits of $$$y_i$$$ such that the resulting number is divisible by $$$60$$$, output "red" (quotes for clarity). Otherwise, output "cyan".

Note: In the first example, there is one rearrangement that yields a number divisible by $$$60$$$, and that is $$$360$$$.In the second example, there are two solutions. One is $$$060$$$ and the second is $$$600$$$.In the third example, there are $$$6$$$ possible rearrangments: $$$025$$$, $$$052$$$, $$$205$$$, $$$250$$$, $$$502$$$, $$$520$$$. None of these numbers is divisible by $$$60$$$.In the fourth example, there are $$$3$$$ rearrangements: $$$228$$$, $$$282$$$, $$$822$$$.In the fifth example, none of the $$$24$$$ rearrangements result in a number divisible by $$$60$$$.In the sixth example, note that $$$000\dots0$$$ is a valid solution.

Sample Inputs:
6
603
006
205
228
1053
0000000000000000000000000000000000000000000000

Sample Outputs:
red
red
cyan
cyan
cyan
red

Solution (Python):
import sys
for _ in range(int(sys.stdin.readline().strip())):
    nums = list(map(int, list(sys.stdin.readline().strip())))
    mod = sum(nums) % 3
    y = min(nums)
    if not y:
        idx = nums.pop(nums.index(0))
        x = min((i % 2 for i in nums))
        if not x and (not mod):
            print('red')
        else:
            print('cyan')
    else:
        print('cyan')

Solution (Python):
def solve(s):
    if int(s) % 3 == 0 and '0' in s:
        x = 0
        for i in s:
            if int(i) % 2 == 0:
                x += 1
            if x == 2:
                return 'red'
    return 'cyan'
n = int(input())
for i in range(n):
    s = input()
    print(solve(s))

Solution (Python):
for i in range(int(input())):
    x = input()
    print('cryeadn'[sum(map(int, x)) % 3 == 0 and x.count('0') and (x.count('0') + x.count('2') + x.count('4') + x.count('6') + x.count('8') > 1)::2])