HDU - 4815(93/600)

一群小动物参观了SYSU的神秘实验室——DeepLab,在这里他们了解到深度学习技术,并见证了一场由聪明的小老虎与实验室创造的智能实体DeepMonkey之间的智力对决。比赛涉及一系列二选一的问题,每个问题都有不同的分数,总分高者获胜。小老虎试图通过策略确保至少获得一定的分数以不输给对手。

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A crowd of little animals is visiting a mysterious laboratory – The Deep Lab of SYSU.

“Are you surprised by the STS (speech to speech) technology of Microsoft Research and the cat face recognition project of Google and academia? Are you curious about what technology is behind those fantastic demos?” asks the director of the Deep Lab. “Deep learning, deep learning!” Little Tiger raises his hand briskly. “Yes, clever boy, that’s deep learning (深度学习/深度神经网络)”, says the director. “However, they are only ‘a piece of cake’. I won’t tell you a top secret that our lab has invented a Deep Monkey (深猴) with more advanced technology. And that guy is as smart as human!”

“Nani ?!” Little Tiger doubts about that as he is the smartest kid in his kindergarten; even so, he is not as smart as human, “how can a monkey be smarter than me? I will challenge him.”

To verify their research achievement, the researchers of the Deep Lab are going to host an intelligence test for Little Tiger and Deep Monkey.

The test is composed of N binary choice questions. And different questions may have different scores according to their difficulties. One can get the corresponding score for a question if he chooses the correct answer; otherwise, he gets nothing. The overall score is counted as the sum of scores one gets from each question. The one with a larger overall score wins; tie happens when they get the same score.

Little Tiger assumes that Deep Monkey will choose the answer randomly as he doesn’t believe the monkey is smart. Now, Little Tiger is wondering “what score should I get at least so that I will not lose in the contest with probability of at least P? ”. As little tiger is a really smart guy, he can evaluate the answer quickly.

You, Deep Monkey, can you work it out? Show your power!�
Input
The first line of input contains a single integer T (1 ≤ T ≤ 10) indicating the number of test cases. Then T test cases follow.

Each test case is composed of two lines. The first line has two numbers N and P separated by a blank. N is an integer, satisfying 1 ≤ N ≤ 40. P is a floating number with at most 3 digits after the decimal point, and is in the range of [0, 1]. The second line has N numbers separated by blanks, which are the scores of each question. The score of each questions is an integer and in the range of [1, 1000]�
Output
For each test case, output only a single line with the answer.
Sample Input
1
3 0.5
1 2 3
Sample Output
3
这玩意是对同一个分数出现的次数进行背包….
次数大于1<

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long 
using namespace std;
int dp[100000];
int tu[41];
main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        double p;
        memset(dp, 0, sizeof(dp));
        cin>>n>>p;
        for(int a=1;a<=n;a++)cin>>tu[a];
        dp[0]=1;
        for(int a=1;a<=n;a++)
        {
            for(int b=99999;b>=0;b--)
            {
                if(b-tu[a]<0)break;
                if(dp[b-tu[a]]>0)
                {
                    if(dp[b]<0)dp[b]=dp[b-tu[a]];
                    else dp[b]+=dp[b-tu[a]];
                }
            }
        }
        int he=0,dan;
        int tt=1ll<<n;
        for(int a=0;a<=99999;a++)
        {
            he+=dp[a];
            if(he*1.0/tt>=p)
            {
                dan=a;
                break;
            }
        }
        cout<<dan<<endl;
    }
}
### 关于HDU - 6609 的题目解析 由于当前未提供具体关于 HDU - 6609 题目的详细描述,以下是基于一般算法竞赛题型可能涉及的内容进行推测和解答。 #### 可能的题目背景 假设该题目属于动态规划类问题(类似于多重背包问题),其核心在于优化资源分配或路径选择。此类问题通常会给出一组物品及其属性(如重量、价值等)以及约束条件(如容量限制)。目标是最优地选取某些物品使得满足特定的目标函数[^2]。 #### 动态转移方程设计 如果此题确实是一个变种的背包问题,则可以采用如下状态定义方法: 设 `dp[i][j]` 表示前 i 种物品,在某种条件下达到 j 值时的最大收益或者最小代价。对于每一种新加入考虑范围内的物体 k ,更新规则可能是这样的形式: ```python for i in range(n): for s in range(V, w[k]-1, -1): dp[s] = max(dp[s], dp[s-w[k]] + v[k]) ``` 这里需要注意边界情况处理以及初始化设置合理值来保证计算准确性。 另外还有一种可能性就是它涉及到组合数学方面知识或者是图论最短路等相关知识点。如果是后者的话那么就需要构建相应的邻接表表示图形结构并通过Dijkstra/Bellman-Ford/Floyd-Warshall等经典算法求解两点间距离等问题了[^4]。 最后按照输出格式要求打印结果字符串"Case #X: Y"[^3]。 #### 示例代码片段 下面展示了一个简单的伪代码框架用于解决上述提到类型的DP问题: ```python def solve(): t=int(input()) res=[] cas=1 while(t>0): n,k=list(map(int,input().split())) # Initialize your data structures here ans=find_min_unhappiness() # Implement function find_min_unhappiness() res.append(f'Case #{cas}: {round(ans)}') cas+=1 t-=1 print("\n".join(res)) solve() ```
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