2

我的朋友为我写了这个脚本来计算一个理论站点所需的建筑材料数量。

它基本上需要2个数字并独立增加它们，直到大数字达到50,000。然后它打印一个这样的列表：

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

我需要对代码进行小幅编辑，以便在打印时将小数字乘以 1.225。我不希望这变得复杂，因为我想保持增量相同。

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()
4

3 回答 3

2

换行：

string += str(getbingint) + ":" + str("%.2f") % getsmallint + ","

和

string += str(getbingint) + ":" + str("%.2f") % (getsmallint*1.225) + ","
于 2010-01-07T11:30:13.670 回答
1

您可以替换string += str(getsmallint)为string += str(getsmallint*1.225)

于 2010-01-07T11:28:07.100 回答
1

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) + ",",
于 2010-01-07T11:43:13.107 回答