Python计算一元函数的N次方多项式
第一种:
from datetime import datetime
#f(x)=a0+a1x+a2x*2+a3x**3+...
def f(a,x):
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p=0
for i in range(0,len(a)):
p=p+a[i]*((x)**i)
print(p)
a=[2,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6]
x=1
begin = datetime.now() # 获取当前datetime
beg_stamp=begin.timestamp()
for i in range(0,100):
t=f(a,x)
end = datetime.now() # 获取当前datetime
end_stamp=end.timestamp() # 把datetime转换为timestamp
print(end_stamp-beg_stamp)
第二种:
from datetime import datetime
#f(x)=a0+x(a1+x(...(an-1+x(an))))
def f(a,x):
p=a[-1]
for i in range(0,len(a)-1):
b=list(reversed(a))
p=b[i+1]+x*p
print(p)
a=[2,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6,9,7,7,8,9,6,3,2,6,8,9,7,7,8,9,6]
x=1
begin = datetime.now() # 获取当前datetime
beg_stamp=begin.timestamp()
for i in range(0,100 ):
t=f(a,x)
end = datetime.now() # 获取当前datetime
end_stamp=end.timestamp() # 把datetime转换为timestamp
print(end_stamp-beg_stamp)
注意:按道理第一种的时间复杂度为n平方,第二种为n
但是做出来的是第一种时间更短 ,不知道为什么???试了很多次
网页标题:Python计算一元函数的N次方多项式
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