c++编程计算x的n次方 c++递归计算x的n次方
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#include
double foo(int n,double x)
{
if(1==n)
{
return x;
}
else
{
if(n%2==0)
return foo(n/2,x*x);
else
return x*foo((n-1)/2,x*x);
}
}
void main()
{
int n;
double x;
cin>>x>>n;
cout<<foo(n,x)<<endl;
}
\u6539\u4e86\u4e00\u4e9b\u9519\u540e\u53ef\u4ee5\u5b9e\u73b0\u6c42x^n\u529f\u80fd\u4e86
运行两次，求出结果。当n自减到零时，跳出循环。返回val。你应该懂得。这很基础。
while（n--）懒人写法，
while(n--)
val*=x;
其等效于
while(n){
val*=x;
n=n-1;
}
当n==0时才退出循环，n不等于0时（如2，3...），循环执行
val*=x;等效于val=val*x;
n--;
if n !=0 继续while内的。。。。
否则 跳出while
#include<bits/stdc++.h>
using namespace std;
int main()
{
int x,y;
cin>>x>>y;
cout<<pow(x,y);
}
x的y次方
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