我试图计算循环内递归调用的复杂性:
Calculate(n,m){
for(i=1; i<n; i++){
if(n==1) return 1;
Calculate(n-1,m);
}
for(j=1; i<m; i++){
Calculate(n,m-1);
if(m==1) return 1;
}
}
这是我做的计算:
C(m,n) = 1 + C(m-1,n) + C(m-2,n) + .... + C(1,n)
+ C(m,n-1) + C(m,n-2) + .... + C(m,1)
是复杂度= 2 ^(m + n)??
答案 0 :(得分:1)
你是对的:要将import csv
import requests
from bs4 import BeautifulSoup
diction_page = ['http://www.bloomberg.com/quote/SPX:IND','http://www.bloomberg.com/quote/CCMP:IND']
for link in diction_page:
res = requests.get(link).text
soup = BeautifulSoup(res,'lxml')
title = soup.select_one('.name').text.strip()
price = soup.select_one('.price').text
print(title,price)
with open('item.csv','a',newline='') as outfile:
writer = csv.writer(outfile)
writer.writerow(["Title","Price"])
writer.writerow([title, price])
或m
减少1,你需要拨2个电话;因为你需要将它们都减少到零,所以你需要经历n
'级别'。
因此,通话次数为O(m+n)
。