区块链 committee
时间: 2023-11-02 22:02:00 浏览: 43
委员会共识机制是一种区块链共识算法,它通过选举少数诚实的节点组成一个委员会来验证和打包区块链上的更新。在基于委员会共识机制的架构中,节点执行本地训练并将本地更新发送给委员会,委员会则负责验证更新的准确性并为其打分。只有符合条件的更新才会被打包到区块链上。下一轮开始时,根据上一轮节点的得分选举新的委员会。委员会成员通过将其数据视为验证集来验证本地更新的准确性,将验证准确性作为分数。综合各委员会成员的分数后,中位数将成为本次更新的分数。委员会共识机制可以提高区块链的稳定性,并避免恶意节点对全局模型的破坏。
相关问题
区块链中dao组织Token模型中都能涉及哪些成员模型
A:在区块链中,DAO组织和Token模型可以涉及以下成员模型:
1. 创始人(Founder):创立DAO组织或Token模型的人或团队。
2. 治理委员会(Governance Committee):负责管理和决策DAO组织或Token模型的委员会。
3. 投资者(Investor):购买Token作为投资的个人或组织。
4. 用户(User):使用Token进行交易或购买商品或服务的个人或组织。
5. 矿工(Miner):为区块链提供计算资源,维护交易记录和验证交易的个人或组织。
6. 节点(Node):运行全节点软件,并参与共识机制的个人或组织。
7. 合作者(Collaborator):与DAO组织或Token模型合作开发或维护的个人或组织。
8. 社区成员(Community Member):对DAO组织或Token模型有兴趣的个人或组织,可以参与讨论和提出建议。
这些成员模型在DAO组织或Token模型的设计和实施中都有不同的角色和职责,共同维护和发展整个生态系统。
用代码解决这个问题The program committee of the school programming contests, which are often held at the Ural State University, is a big, joyful, and united team. In fact, they are so united that the time spent together at the university is not enough for them, so they often visit each other at their homes. In addition, they are quite athletic and like walking. Once the guardian of the traditions of the sports programming at the Ural State University decided that the members of the program committee spent too much time walking from home to home. They could have spent that time inventing and preparing new problems instead. To prove that, he wanted to calculate the average distance that the members of the program committee walked when they visited each other. The guardian took a map of Yekaterinburg, marked the houses of all the members of the program committee there, and wrote down their coordinates. However, there were so many coordinates that he wasn't able to solve that problem and asked for your help. The city of Yekaterinburg is a rectangle with the sides parallel to the coordinate axes. All the streets stretch from east to west or from north to south through the whole city, from one end to the other. The house of each member of the program committee is located strictly at the intersection of two orthogonal streets. It is known that all the members of the program committee walk only along the streets, because it is more pleasant to walk on sidewalks than on small courtyard paths. Of course, when walking from one house to another, they always choose the shortest way. All the members of the program committee visit each other equally often. Input The first line contains the number n of members of the program committee (2 ≤ n ≤ 105). The i-th of the following n lines contains space-separated coordinates xi, yi of the house of the i-th member of the program committee (1 ≤ xi, yi ≤ 106). All coordinates are integers. Output Output the average distance, rounded down to an integer, that a member of the program committee walks from his house to the house of his colleague.
Sure, I can help you with that. Here's one possible solution in Python:
```python
import math
n = int(input())
x_coords = []
y_coords = []
for i in range(n):
x, y = map(int, input().split())
x_coords.append(x)
y_coords.append(y)
x_coords.sort()
y_coords.sort()
x_median = x_coords[n // 2]
y_median = y_coords[n // 2]
total_distance = 0
for i in range(n):
total_distance += abs(x_coords[i] - x_median) + abs(y_coords[i] - y_median)
average_distance = total_distance // n
print(average_distance)
```
The idea here is to find the median x-coordinate and the median y-coordinate of all the houses, since these will be the coordinates of the "center" of the city. Then, we calculate the total distance each member of the program committee will have to walk to get to their colleague's house, by adding up the Manhattan distances between their own coordinate and the median coordinates. Finally, we divide the total distance by the number of committee members to get the average distance, rounded down to an integer. Note that we don't actually need to calculate the square root of the distances, since we only care about the integer value.
I hope this helps! Let me know if you have any questions.