1 solutions
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3
坑人的递归题!!!
题目所描述的代码:
#include<iostream> #include<cmath> using namespace std; double f(double x,double n) { if(n<=1) { return sqrt(1+x); } else { return sqrt(n+f(x,n-1)); } } int main() { double a,b; cin>>a>>b; printf("%.2lf",f(a,b)); return 0; }
实际上的代码(我查网上了):
根本就不是平方根!!!
是除法!!!
我就说怎么样例对不上!!!
#include<iostream> #include<cmath> using namespace std; double f(double x,double n)//递归函数 { if(n<=1)//递归边界 { return x/(1+x);//结束递归 } else { return x/(n+f(x,n-1));//递归 } } int main() { double a,b; cin>>a>>b;//输入 printf("%.2lf",f(a,b));//保留两位(题目也没说)!!! return 0;//完结散花 }
- 1
Information
- ID
- 653
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 33
- Accepted
- 16
- Uploaded By