计算机导论课第二次作业 (Past)
10205102406 edited 3 年，6 月前
This is a past version of blog 计算机导论课第二次作业
1、略
2、
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def sqr(x):
x = abs(x)
res = 0
for i in range(x):
res += x
return res
def cube(x):
t = sqr(x)
res = 0
for i in range(abs(x)):
res += t
if x > 0:
return res
else:
return -res
def quad(x):
if x == 0:
return 0
else:
return int(sqr(x)/(1/sqr(x)))
print(sqr(5), sqr(-5))
print(cube(5), cube(-5))
print(quad(5), quad(-5))
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3、
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def factor(n):
res = n
for i in range(2,n):
res *= i
return res
def combination(n, k):
return int(factor(n)/(factor(n-k)*factor(k)))
print(factor(4))
print(combination(4, 2))
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4、因为3/x不是整除，会有小数点，x太大的时候因为浮点数精度不足就把3/x当成0了，所以3/x+1就是1啦，但是x不够大的时候3/x就没有被当做0，还有一点点，所以3/x+1就大于1，而不是等于1啦。
5、
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print(2**10)
print(2**20)
print(2**30)
print(2**40)
print(2**50)
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6、
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def square_root_1():
c = 10
i = 0
g = 0
while (g+1)**2 < c:
g+=1
while (abs(g*g-c) > 0.0001):
g += 0.00001
i = i+1
print("% d:g = %.5f" % (i,g))
square_root_1()
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7、
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def square_root_2():
i = 0
c = 0.9
m_max = c if c>1 else 1
m_min = 0
g = (m_min + m_max)/2
while(abs(gg-c) > 0.00000000001):
if(gg<c):
m_min = g
else:
m_max = g
g = (m_min + m_max)/2
i = i+1
print(“%d:%.13f” % (i,g))
square_root_2()
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8、
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x = 1
c = 0
i = 1.5
while abs(x-i) > 0.112:
x = i
if x3+x2-10 > 0:
i = x - 0.5 (c+2)
else:
i = x + 0.5 ** (c+2)
c+=1
print(“二分法结果为%.12f，共迭代%d次”%(i,c))
x = 1
c = 0
i = 1.5
while abs(x-i) > 0.112:
x = i
i = x-(x3+x2-10)/(3*x2+2*x)
c+=1
print(“牛顿法结果为%.12f，共迭代%d次”%(i,c))
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9、
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A = [[1,2,6],
[2,4,12]]
D = A[0][0]A[1][1] - A[0][1]A[1][0]
Dx = A[0][2]A[1][1] - A[0][1]A[1][2]
Dy = A[0][0]A[1][2] - A[0][2]A[1][0]
if D == 0:
if Dx == 0 and Dy == 0:
print(“方程有无穷解”)
else:
print(“方程无解”)
else:
print(“方程有解:x = %f, y = %f” %(Dx/D, Dy/D))
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10、
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s = “I am a Chinese “
res = []
word = “”
for i in s:
if i == ‘ ‘:
if word:
res.insert(0, word)
word = “”
else:
word += i
if word:
res.insert(0, word)
print(res)
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11、
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s = “ABABA”
if s == s[::-1]:
print(True)
else:
print(False)
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12、略
编辑者：姜博远