ZOJ3958-Cooking Competition

本文介绍了一个基于《小林家的龙女仆》中的一场烹饪比赛的评分系统。该系统通过模拟不同类型的反馈来确定比赛的胜者,并提供了一段示例代码实现。

Cooking Competition

Time Limit: 1 Second      Memory Limit: 65536 KB

"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+10
2Tohru cooks better0+1
3Both of them are good at cooking+1+1
4Both 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 a1a2, ... , 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.


Author: WENG, Caizhi
Source: The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple


题意:有四种给分方式,问谁的分数高

解题思路:模拟


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[6] = { 0,1,0,1,-1 };
int b[6] = { 0,0,1,1,-1 };

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int x,n;
		int ans1 = 0, ans2 = 0;
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			ans1 += a[x];
			ans2 += b[x];
		}
		if (ans1 > ans2) printf("Kobayashi\n");
		else if (ans1 < ans2) printf("Tohru\n");
		else printf("Draw\n");
	}
	return 0;
}

先展示下效果 https://pan.quark.cn/s/a4b39357ea24 遗传算法 - 简书 遗传算法的理论是根据达尔文进化论而设计出来的算法: 人类是朝着好的方向(最优解)进化,进化过程中,会自动选择优良基因,淘汰劣等基因。 遗传算法(英语:genetic algorithm (GA) )是计算数学中用于解决最佳化的搜索算法,是进化算法的一种。 进化算法最初是借鉴了进化生物学中的一些现象而发展起来的,这些现象包括遗传、突变、自然选择、杂交等。 搜索算法的共同特征为: 首先组成一组候选解 依据某些适应性条件测算这些候选解的适应度 根据适应度保留某些候选解,放弃其他候选解 对保留的候选解进行某些操作,生成新的候选解 遗传算法流程 遗传算法的一般步骤 my_fitness函数 评估每条染色体所对应个体的适应度 升序排列适应度评估值,选出 前 parent_number 个 个体作为 待选 parent 种群(适应度函数的值越小越好) 从 待选 parent 种群 中随机选择 2 个个体作为父方和母方。 抽取父母双方的染色体,进行交叉,产生 2 个子代。 (交叉概率) 对子代(parent + 生成的 child)的染色体进行变异。 (变异概率) 重复3,4,5步骤,直到新种群(parentnumber + childnumber)的产生。 循环以上步骤直至找到满意的解。 名词解释 交叉概率:两个个体进行交配的概率。 例如,交配概率为0.8,则80%的“夫妻”会生育后代。 变异概率:所有的基因中发生变异的占总体的比例。 GA函数 适应度函数 适应度函数由解决的问题决定。 举一个平方和的例子。 简单的平方和问题 求函数的最小值,其中每个变量的取值区间都是 [-1, ...
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