怎么用pathon写出两个时间变量,相差1秒
时间: 2023-05-23 16:04:04 浏览: 53
可以使用datetime模块来实现:
```python
import datetime
time1 = datetime.datetime.now()
time2 = time1 + datetime.timedelta(seconds=1)
print(time1)
print(time2)
```
输出为:
```
2022-02-14 14:25:20.752873
2022-02-14 14:25:21.752873
```
相关问题
用pathon写出dijikstra
好的,这是一个基于Python的Dijkstra算法示例代码:
```python
import heapq
def dijkstra(graph, start):
# 初始化距离字典和堆
distances = {vertex: float('infinity') for vertex in graph}
distances[start] = 0
heap = [(0, start)]
while heap:
# 弹出堆中最小距离的节点
current_distance, current_vertex = heapq.heappop(heap)
# 如果当前距离已经大于已知最短距离,跳过
if current_distance > distances[current_vertex]:
continue
# 遍历当前节点的邻居节点,更新距离
for neighbor, weight in graph[current_vertex].items():
distance = current_distance + weight
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(heap, (distance, neighbor))
return distances
```
其中,`graph`是一个字典类型,表示图的邻接表,`start`是起点节点。示例代码中使用了堆(heap)来优化算法时间复杂度。
用pathon写出点菜系统
好的,以下是一个简单的点菜系统的 Python 代码:
```python
# 初始化菜单
menu = {
"炒饭": 15,
"炒面": 12,
"鱼香肉丝": 20,
"宫保鸡丁": 18,
"红烧肉": 22
}
# 初始化订单
order = {}
# 展示菜单
print("欢迎光临餐厅!")
print("这是我们的菜单:")
for dish, price in menu.items():
print(f"{dish}:{price}元")
# 开始点菜
while True:
dish_name = input("请输入您要点的菜名(输入q结束点菜):")
if dish_name == "q":
break
elif dish_name in menu:
dish_price = menu[dish_name]
dish_num = int(input(f"请输入{dish_name}的数量:"))
order[dish_name] = dish_num
print(f"{dish_name}已加入您的订单,单价{dish_price}元,数量{dish_num}份。")
else:
print("很抱歉,我们没有这道菜。")
# 展示订单
if order:
print("您的订单如下:")
total_price = 0
for dish, num in order.items():
price = menu[dish] * num
print(f"{dish}:{num}份,共{price}元")
total_price += price
print(f"总价为{total_price}元。")
else:
print("您没有点任何菜。")
```
这个点菜系统会先展示菜单,然后循环让用户输入要点的菜名和数量,并将点的菜添加到一个订单中。最后,系统会展示订单和总价。
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