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假设我有一个函数int fun1();,它可以返回一个随机数 x1 满足0<x1<m。
int fun1();
0<x1<m
如何利用该函数fun1()生成另一个函数int fun2(),使其返回一个满足 x2 的随机数0<x2<n?
fun1()
int fun2()
0<x2<n
您需要编写如下内容:
return fun1() * n / m;
I have a requirement to pass an IEnumerable<Object> to an MVC view and have the view cast this to a strongly typed object and display it according
IEnumerable<Object>