double Hsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);这个表达式求的结果是无穷大,什么原因?
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doubleHsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);这个表达式求的结果是无穷大
double Hsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);这个表达式求的结果是无穷大,什么原因?
double Hsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);
这个表达式求的结果是无穷大,什么原因?
double Hsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);这个表达式求的结果是无穷大,什么原因?
1/56*Math.Log10(0.1) = 0
c++是这样,你的是C#吧,可能情况类似.
==>
1/56.*.
double Hsmax=0.594+2.29*Math.Pow((-Math.Log10(1-Math.Exp(1/56*Math.Log10(0.1)))),0.722);这个表达式求的结果是无穷大,什么原因?
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double ** p=new double*
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【c++】现有三个函数(1)double Add(double a) (2)double Add(double a,double b) (3)……如题,(3) int Add(double a,double b,double c=0)现在对Add函数进行调用,编译出现错误的是A.Add(3) B.Add(3.3,3.4) C.Add(2.3,3.1,3.2) D.Add(3
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