题解 | #加减# B.Austin Love Math

$f(x)=\frac{x}{\sqrt{x^2+1}}f(x)\; =\; \backslash frac\{x\}\{\backslash sqrt\{x^2+1\}\}$

$f(x)=\frac{1}{\sqrt{1+\frac{1}{x^2}}}f(x)\; =\backslash frac\{1\}\{\backslash sqrt\{1\; +\; \backslash frac\{1\}\{x^2\}\}\}$

$\frac{1^2}{f(x{)}^2}=1+\frac{1}{x^2}\backslash frac\{1^2\}\{f(x)^2\}=1+\backslash frac\{1\}\{x^2\}$

$f(f(x))=\frac{1}{\sqrt{1+\frac{1^2}{f(x{)}^2}}}f(f(x))\; =\; \backslash frac\{1\}\{\backslash sqrt\{1+\backslash frac\{1^2\}\{f(x)^2\}\}\}$

$f_2(x)=f(f(x))=\frac{1}{\sqrt{1+1+\frac{1}{x^2}}}f\_2(x)=f(f(x))\; =\backslash frac\{1\}\{\backslash sqrt\{1+1+\backslash frac\{1\}\{x^2\}\}\}$

同理计算$f_3(x)f\_3(x)$

$f_2(x)=\frac{1}{\sqrt{2+\frac{1}{x^2}}},f_3(x)=\frac{1}{\sqrt{3+\frac{1}{x^2}}}f\_2(x)\; =\backslash frac\{1\}\{\backslash sqrt\{2+\backslash frac\{1\}\{x^2\}\}\}\; ,f\_3(x)=\backslash frac\{1\}\{\backslash sqrt\{3+\backslash frac\{1\}\{x^2\}\}\}$

..... 则$f_n=\frac{1}{\sqrt{n+\frac{1}{x^2}}}f\_n=\backslash frac\{1\}\{\backslash sqrt\{n+\backslash frac\{1\}\{x^2\}\}\}$

注意$f_1(x)\mathrm{\ne }f(x),f_0(x)=f(x)f\_1(x)\; \backslash neq\; f(x)\; ,f\_0(x)=f(x)$

double calc(int x , int t){
	return 1.00/sqrt( t + 1.00/x/x);
}
void solve(){
	int x,t;cin>>x>>t;
	cout<<calc(x,t + 1)<<endl;
	
}