Python基础数学
Python基础数学
提问于 2010-01-07 19:24:07
我的朋友为我写了这个脚本，用来计算一个理论站点所需的建筑材料的数量。
它基本上取2个数字，然后独立地递增，直到大数字达到50,000。然后，它会打印出如下列表：
20000:6.40,21000:6.61,22000:6.82,23000:7.03,24000:7.24,25000:7.45,26000:7.66,27000:7.87,28000:8.08,29000:8.29,30000:8.50,31000:8.71,32000:8.92,33000:9.13,34000:9.34,35000:9.55,36000:9.76,37000:9.97,38000:10.18,39000:10.39,40000:10.60,41000:10.81,42000:11.02,43000:11.23,44000:11.44,45000:11.65,46000:11.86,47000:12.07,48000:12.28,49000:12.49,50000:12.70我需要对代码做一个小的编辑，以便在打印时将小数字乘以1.225。我不想让它变得复杂，因为我想保持增量不变。
getbingint = input("Enter big start value: ")
getbiginc = input("Enter big increment value: ")
getsmallint = input("Enter small start value: ")
getsmalinc = input("Enter small increment value: ")
getbigend = input("Enter big end value: ")
string = ""
while getbingint <= getbigend:
string += str(getbingint) + ":" + str("%.2f") % getsmallint + ","
getbingint += getbiginc
getsmallint += getsmalinc
print string
raw_input()回答 3
Stack Overflow用户
回答已采纳
发布于 2010-01-07 19:30:14
替换该行：
string += str(getbingint) + ":" + str("%.2f") % getsmallint + "," 使用
string += str(getbingint) + ":" + str("%.2f") % (getsmallint*1.225) + "," Stack Overflow用户
发布于 2010-01-07 19:28:07
您可以用string += str(getsmallint*1.225)替换string += str(getsmallint)
Stack Overflow用户
发布于 2010-01-07 19:43:13
这是另一个版本
getbingint = input("Enter big start value: ")
getbiginc = input("Enter big increment value: ")
getsmallint = input("Enter small start value: ")
getsmalinc = input("Enter small increment value: ")
getbigend = input("Enter big end value: ")
for i in range(getbingint,getbigend,getbiginc+1):
getsmallint += getsmalinc
print str(i) +":"+ str("%.2f") % (getsmallint*1.225) + ",",页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/2019850
复制相关文章
相似问题
腾讯云开发者
