Red's deleted number

Classification of discussion sum%3 == 0, == 1 == 2

among sum%3 == 0 Can be applied to =1, be equal to 2 The situation of

Be careful n = 1 Etc Judgment of the situation

And when the condition for Xiaozi to win is equal to %2==1 When the sum is not equal to >2

The important method is to enumerate the special cases , Enumerate from quantity , There are only three equivalence classes

// Problem:  Xiao Hong's deleted number 
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11251/C
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-06-10 20:17:47
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
    ll ans = 0;
    while (y) {
    	if (y & 1)
    		ans = (ans + x) %mod;
    	y >>= 1;
    	x = (x + x) %mod;
	}
	return ans;
}
inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a;
		a = a * a;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
// inline ll inv (ll a) {
	// return qmi(a, mod - 2);
// }

int num[5];
int n, sum;
string s;

void solve() {
	cin >> s;
	for (auto t : s) {
		int c = t - '0';
		if (c == 0) continue;
		
		n ++;
		c%=3;
		num[c] ++;
		sum += t -'0';
	}
	
	if (sum % 3 == 0) {
		if (num[0] == 0 || n <= 1) {
			puts("yukari");
			return;
		}
		num[0] --;
		n --;
		if (num[0] == 0) {
			if (n == 1) {
				puts("kou");
				return;
			}
			if (num[1] != num[2] && abs(num[1] - num[2]) > 1
			|| (num[1] == num[2] && num[1]%2)) 
			{
				puts("yukari") ;
			}
			else puts("kou");
		}
		else {
			if (num[1] != num[2])
				puts("yukari");
			else 
			{
				 puts("kou");
			}
		}
	}
	else if (sum % 3 == 1) {
		if (num[1] == 0 || n <= 1) {
			puts("yukari");
			return;
		}
		num[1] --;
		n --;
		if (num[0] == 0) {
			if (n == 1) {
				puts("kou");
				return;
			}
			if (num[1] != num[2] && abs(num[1] - num[2]) > 1
			|| (num[1] == num[2] && num[1]%2)) 
			{
				puts("yukari") ;
			}
			else puts("kou");
		}
		else {
			if (num[1] != num[2])
				puts("yukari");
			else 
			{
				 puts("kou");
			}
		}
	}
	else if (sum % 3==2) {
		if (num[2] == 0 || n <= 1) {
			puts("yukari");
			return;
		}
		num[2] --;
		n --;
		if (num[0] == 0) {
			if (n == 1) {
				puts("kou");
				return;
			}
			if (num[1] != num[2] && abs(num[1] - num[2]) > 1
			|| (num[1] == num[2] && num[1]%2)) 
			{
				puts("yukari") ;
			}
			else puts("kou");
		}
		else {
			if (num[1] != num[2])
				puts("yukari");
			else 
			{
				puts("kou");
			}
		}
	}
	
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}