1266A - Competitive Programmer - CodeForces Solution

chinese remainder theorem math *1000

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Python Code: 

tests = int(input())

for _ in range(tests):
    n = input()
    if n.count('0') == 0:
        print("cyan")
        continue
    par = 0
    sum = 0
    for num in n.replace('0','',1):
        sum += int(num)
        if(int(num) % 2 == 0):
            par += 1

    if par == 0:
        print("cyan")
        continue

    if sum % 3 != 0:
        print("cyan")
        continue
    print("red")

C++ Code: 

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll gcd(ll a,ll b)
{
    ll R;
    while ((a % b) > 0)
    {
        R = a % b;
        a = b;
        b = R;
    }
    return b;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
//    vector <bool>prim(100000000);
//    for (ll i=2 ;i<100000 ;i++)
//    {
//        if (!prim[i])
//        {
//            ll k=2;
//            while (k*i<=100000)
//            {
//                prim[k*i]=1;
//                k++;
//            }
//        }
//    }
    int tc=1;
    cin >> tc;
    int kkkkk=tc;
    while(tc--)
    {
//        cout <<kkkkk-tc<<"------->\n";
        string s;
        cin >> s;
        int z=0,t=0;
        int sum=0;
        for (int i=0 ;i<s.size() ;i++)
        {
            sum+=s[i]-'0';
            if (s[i]=='0')
                z++;
            else if((s[i]-'0')%2==0)
                t=1;
        }
        if ((z && (t || z>1) && sum%3==0) || sum==0)
            cout << "red\n";
        else
            cout << "cyan\n";
    }
    return 0;
}


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