C++(g++ 7.5.0) 解法, 执行用时: 6ms, 内存消耗: 652K, 提交时间: 2022-11-22 20:17:18

#include<bits/stdc++.h>
using namespace std;
int tong[3];

int main(){

  int n,x,i,sum = 0;
  string s;
  cin >> s;
  for(i=0;i<s.length();++i){
    if(s[i]-'0') tong[(s[i]-'0')%3]++;
    sum += s[i]-'0';
  }
  if(!tong[sum%3]) {
    cout << "yukari";
    return 0;
  }
  tong[sum % 3]--;
  if(tong[1] == tong[2] && tong[0]) cout<<"kou";
  else cout<<"yukari";
}

Python3 解法, 执行用时: 86ms, 内存消耗: 5480K, 提交时间: 2022-09-21 09:19:35

from collections import Counter
s = input().strip()
ns = [int(c)%3 for c in s if c != '0']

cnt = Counter(ns)
Sum = sum(ns)

def main():
    if cnt[Sum % 3] == 0:
        return 'yukari'
    cnt[Sum % 3] -= 1
    if cnt[1]==cnt[2] and cnt[0]>0:
        return 'kou'
    else:
        return 'yukari'
    

print(main())

C++ 解法, 执行用时: 5ms, 内存消耗: 524K, 提交时间: 2022-06-14 22:59:11

#include<cstdio>
int x,n[3],p;
char s[100005];
int main(){
	scanf("%s",s);
	for(int i=0;s[i];i++){
		p=(p*10+s[i]-'0')%3;
		if(s[i]=='0') continue;
		++n[(s[i]-'0')%3];
	}
	if(!n[p]) goto B;
	--n[p];
	if(n[1]==n[2]&&n[0]) goto A;
	goto B;
A:
	puts("kou");
	return 0;
B:
	puts("yukari");
	return 0;
}