补全上述代码
时间: 2023-07-03 07:22:19 浏览: 42
好的,下面是完整的代码:
```python
import openai
import requests
# 设置 CSDN API 访问密钥
openai.api_key = "YOUR_API_KEY"
# 输入对话信息
prompt = "你好,我是人工智能助手,请问有什么需要帮助的吗?"
# 调用 CSDN 的 GPT-3 模型生成回复
response = openai.Completion.create(
engine="davinci",
prompt=prompt,
max_tokens=60,
n=1,
stop=None,
temperature=0.7,
)
# 输出回复
print(response.choices[0].text.strip())
```
其中,需要将 `YOUR_API_KEY` 替换为你的 CSDN API 访问密钥。此外,你可以根据自己的需要修改 `prompt`、`max_tokens`、`temperature` 等参数。
相关问题
请补全上述代码
非常抱歉,我发现我的回答中有些地方没有完全补全代码。以下是完整的代码:
```
import gym
import math
import random
import torch
import torch.nn as nn
import torch.nn.functional as F
import torch.optim as optim
from Q_net import Net
class Agent(object):
def __init__(self, **kwargs):
for key, value in kwargs.items():
setattr(self, key, value)
self.eval_net = Net(self.state_space_dim, self.action_space_dim)
self.optimizer = optim.Adam(self.eval_net.parameters(), lr=self.lr)
self.buffer = []
self.steps = 0
def act(self, s0):
self.steps += 1
epsi = self.epsi_low + (self.epsi_high - self.epsi_low) * (math.exp(-1.0 * self.steps / self.decay))
if random.random() < epsi:
a0 = random.randrange(self.action_space_dim)
else:
s0 = torch.tensor(s0, dtype=torch.float).view(1, -1)
# 根据状态选择最佳动作
with torch.no_grad():
q_values = self.eval_net(s0)
a0 = q_values.argmax(dim=1)
return a0
def put(self, *transition):
if len(self.buffer) == self.capacity:
self.buffer.pop(0)
self.buffer.append(transition)
def learn(self):
if (len(self.buffer)) < self.batch_size:
return
samples = random.sample(self.buffer, self.batch_size)
s0, a0, r1, s1 = zip(*samples)
s0 = torch.tensor(s0, dtype=torch.float)
a0 = torch.tensor(a0, dtype=torch.long).view(self.batch_size, -1)
r1 = torch.tensor(r1, dtype=torch.float).view(self.batch_size, -1)
s1 = torch.tensor(s1, dtype=torch.float)
loss_fn = nn.MSELoss()
# 定义损失函数
q_next = self.eval_net(s1).max(dim=1, keepdim=True)[0].detach()
q_target = r1 + self.gamma * q_next
q_eval = self.eval_net(s0).gather(1, a0)
loss = loss_fn(q_eval, q_target)
self.optimizer.zero_grad()
loss.backward()
self.optimizer.step()
```
希望这次的回答能够帮助到您,如有疑问请随时提问。
补全上述代码解决约瑟夫环问题
好的,这里是补全后的代码。这段代码实现了约瑟夫环问题的求解,其中使用了循环链表。
```c
#include <stdio.h>
#include <stdlib.h>
struct student {
int num;
struct student *next;
};
struct student *createlink(int);
struct student *createnode(int);
void del(struct student *, struct student *);
void output(struct student *);
void josephus(int, int, int);
int main() {
int n = 5, m = 3, k = 2;
josephus(n, m, k);
return 0;
}
void josephus(int n, int m, int k) {
struct student *head = createlink(n);
struct student *p = head->next;
int i;
for (i = 1; i < k; i++) {
head = head->next;
p = p->next;
}
while (p->next != p) {
for (i = 1; i < m; i++) {
head = head->next;
p = p->next;
}
printf("出队编号:%d\n", p->num);
head->next = p->next;
free(p);
p = head->next;
}
printf("最后一个出队编号:%d\n", p->num);
free(p);
}
struct student *createlink(int n) {
struct student *head, *p1, *p2;
int i;
for (i = 1; i <= n; i++) {
if (i == 1)
head = p1 = createnode(i);
else {
p2 = createnode(i);
p1->next = p2;
p1 = p2;
}
}
p1->next = head;
return head;
}
struct student *createnode(int num) {
struct student *p;
p = (struct student *)malloc(sizeof(struct student));
p->num = num;
p->next = NULL;
return p;
}
```
在`main`函数中,调用`josephus`函数解决约瑟夫环问题。`josephus`函数接受三个参数:n表示队列的长度,m表示每次出队的间隔,k表示从第k个人开始出队。
在函数内部,首先创建循环链表,然后根据k找到第一个出队的人。接着进入循环,每次找到第m个人,并将其出队,直到只剩下最后一个人为止。最后释放内存并输出结果。
需要注意的是,由于循环链表的特殊性,每次找到第m个人时需要同时更新头结点和当前结点。
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