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@@ -0,0 +1,51 @@
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+#count sqrt via newton iteration
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+'''
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+算法：
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+对于曲线f(x)=0,如果一阶导函数f_nd(x)存在，可以利用牛顿迭代来快速求得近似解。
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+对于初次预估近似解 x_0, 有切线方程：y-f(x_0)=(x-x_0)*f_nd(x_0)
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+真实解对应于 y = 0, x = x_n
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+(x_n - x_(n-1))f_nd(x_0) + f(x_(n-1))=0
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+x_n = x_(n-1) - f(x_(n-1))/f_nd(x_(n-1))
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+当 x_n - x(n-1) < 指定误差时，我们认为x_n 是可以接受的近似值
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+对于求a 的 平方根而言，则方程如下：
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+f(x) = x**2 - a
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+f_nd(x) = 2x
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+x_0 = a
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+x = x_(n-1)-((x_(n-1))**2-a)/2(x_(n-1))
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+'''
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+
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+import optparse
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+import sys
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+Epsilon = 1e-15
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+
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+
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+def Newton_sqrt(c):
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+
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+ t = c
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+ while abs(t - c/t)>Epsilon:
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+
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+ t = (c/t + t)/2.0
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+
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+
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+ print('the sqrt of {0} is {1:.5f}'.format(c,t))
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+
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+
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+
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+def main():
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+ '''
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+ 帮助信息
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+ 获取参数
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+ '''
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+
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+
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+ parser = optparse.OptionParser("""\
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+usage: %prog [options] infile outfile
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+tell you the sqrt of your input number""")
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+
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+ parser.add_option("-r", "--sqrt", dest="sqrt",
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+ help=("get the sqrt"))
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+ opts, args = parser.parse_args()
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+ Newton_sqrt(float(opts.sqrt))
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+
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+
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+main()
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