Cooking Competition

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"Miss Kobayashi's Dragon Maid" is a Japanese manga series written and illustrated by Coolkyoushinja. An anime television series produced by Kyoto Animation aired in Japan between January and April 2017.


In episode 8, two main characters, Kobayashi and Tohru, challenged each other to a cook-off to decide who would make a lunchbox for Kanna's field trip. In order to decide who is the winner, they asked n people to taste their food, and changed their scores according to the feedback given by those people.


There are only four types of feedback. The types of feedback and the changes of score are given in the following table.


Type Feedback Score Change
(Kobayashi) Score Change
(Tohru)
1 Kobayashi cooks better +1 0
2 Tohru cooks better 0 +1
3 Both of them are good at cooking +1 +1
4 Both of them are bad at cooking -1 -1
Given the types of the feedback of these n people, can you find out the winner of the cooking competition (given that the initial score of Kobayashi and Tohru are both 0)?


Input


There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 100), indicating the number of test cases. For each test case:


The first line contains an integer n (1 ≤ n ≤ 20), its meaning is shown above.


The next line contains n integers a1, a2, ... , an (1 ≤ ai ≤ 4), indicating the types of the feedback given by these n people.


Output


For each test case output one line. If Kobayashi gets a higher score, output "Kobayashi" (without the quotes). If Tohru gets a higher score, output "Tohru" (without the quotes). If Kobayashi's score is equal to that of Tohru's, output "Draw" (without the quotes).


Sample Input


2
3
1 2 1
2
3 4
Sample Output


Kobayashi
Draw
Hint


For the first test case, Kobayashi gets 1 + 0 + 1 = 2 points, while Tohru gets 0 + 1 + 0 = 1 point. So the winner is Kobayashi.


For the second test case, Kobayashi gets 1 - 1 = 0 point, while Tohru gets 1 - 1 = 0 point. So it's a draw.


S

题意：

有四中结果，编号为1 2 3 4,其中 1代表给k加1分，2代表给T加一分，3代表给K T均加1分，4为均减1分，然后输入数据，判断哪个人得的分数多，输出它！

思路：

使用几组if选择情况即可！

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t,n,m,i;
	cin>>t;
	while(t--)
	{
		cin>>n;
		int a=0,b=0;
		for(i=0;i<n;i++)
		{
			cin>>m;
			if(m==1)
			{
				a++;
			}
			else if(m==2)
			{
				b++;
			}
			else if(m==3)
			{
				a++;
				b++;
			}
			else
			{
				a--;
				b--;
			}
			
		}
		if(a>b)
				cout<<"Kobayashi"<<endl;
			else if(a<b)
				cout<<"Tohru"<<endl;
			else
				cout<<"Draw"<<endl;
	}
	
	
	return 0;
}

水题！！！下次做水题争取8分钟解决！！！

本文标签： Cookingcompetition