第十一届浙江省程序设计竞赛

本文介绍了第十一届浙江省程序设计竞赛的多个题目,包括Pokemon Master、Problem Arrangement、Talented Chef等。每个题目提供了输入输出格式和样例,涉及比较权重、计算期望值、时间规划等解题思路。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >


A.Pokemon Master

Calem and Serena are pokemon masters. One day they decided to have a pokemon battle practice before Pokemon World Championships. Each of them has some pokemons in each's team. To make the battle more interesting, they decided to use a special rule to determine the winner: the team with heavier total weight will win the battle!

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N and M (1 <= NM <= 6), which describes that Calem has N pokemons and Serena has M pokemons.

The second line contains N integers indicating the weight of Calem's pokemons. The third line contains M integers indicating the weight of Serena's pokemons. All pokemons' weight are in the range of [1, 2094] pounds.

Output

For each test case, output "Calem" if Calem's team will win the battle, or "Serena" if Serena's team will win. If the two team have the same total weight, output "Draw" instead.

Sample Input
1
6 6
13 220 199 188 269 1014
101 176 130 220 881 396
Sample Output
Serena

水题,直接比大小即可


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
	int num,t,i,j,sum1,sum2,n,m;
	cin>>t;
	while(t--) {
		sum1=sum2=0;
		cin>>n>>m;
		for(i=1;i<=n;i++) {
			cin>>num;
			sum1+=num;
		}
		for(i=1;i<=m;i++) {
			cin>>num;
			sum2+=num;
		}
		if(sum1>sum2) cout<<"Calem"<<endl;
		else if(sum1<sum2) cout<<"Serena"<<endl;
		else cout<<"Draw"<<endl;
	}
	return 0;
} 


B.Problem Arrangement


The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order of the problems. As we know, the arrangement will have a great effect on the result of the contest. For example, it will take more time to finish the first problem if the easiest problem hides in the middle of the problem list.

There are N problems in the contest. Certainly, it's not interesting if the problems are sorted in the order of increasing difficulty. Edward decides to arrange the problems in a different way. After a careful study, he found out that the i-th problem placed in the j-th position will add Pij points of "interesting value" to the contest.

Edward wrote a program which can generate a random permutation of the problems. If the total interesting value of a permutation is larger than or equal to M points, the permutation is acceptable. Edward wants to know the expected times of generation needed to obtain the first acceptable permutation.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each

评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值