2017.07.16小组赛题目D

D - Cooking Competition

 

"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.

TypeFeedbackScore Change
(Kobayashi)Score Change
(Tohru)1Kobayashi cooks better+102Tohru cooks better0+13Both of them are good at cooking+1+14Both 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, ... , a_n (1 ≤ a_i ≤ 4), indicating the types of the feedback given by these n people.

<h4< dd="">

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).

<h4< dd="">

Sample Input

231 2 123 4

Sample Output

KobayashiDraw

 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.、

本文大体的意思是：有这么两个人，在以下的评分人下面，谁的分数高就是谁赢

C语言代码：

#include<stdio.h>
int a[5]={0,1,0,1,-1};
int b[5]={0,0,1,1,-1};
 int main()
 {
  int n;
  int digit[105];
  while(scanf("%d",&n)!=EOF)
  {
  int i,m,j;
  int sum1,sum2;
  for(i=0;i<n;i++)
  {
  scanf("%d",&m);
  sum1=0;
sum2=0;
  for(j=0;j<m;j++)
  {
  scanf("%d",&digit[j]);
     sum1=sum1+a[digit[j]];
sum2=sum2+b[digit[j]];
  }
   if(sum1==sum2)printf("Draw\n");
      else if(sum1>sum2)printf("Kobayashi\n");
 else if(sum1<sum2)printf("Tohru\n");
  }
  } 
 return 0;
 }