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Brief Theory : Wiki Link
Code 1 : Basic (Worst)
#include<bits/stdc++.h>
using namespace std;
typedef long long l;
typedef long double f;
int main(){
l t,i,n;
cout<<"No of testcases: ";
cin>>t;
while(t--){
cout<<"Decimal No of precision :";
cin>>n;
n=ceil(sqrt(pow(10,n+1)));
l d=0;
f a=1,b=1,r;
while(d<=n){
d++;
if(d%2)
a=b-1/pow(2*d+1,2);
else b=a+1/pow(2*d+1,2);
}
r=(n%2) ? a:b;
cout<<"Catalan's constant for "<<n<<"th iteration: "<<setprecision(50)<<r<<endl;
}
return 0;
}
Code 2 (Better)
#include<bits/stdc++.h>
using namespace std;
typedef long long l;
typedef long double f;
int main(){
l t,i,n;
cout<<"No of testcases: ";
cin>>t;
while(t--){
cout<<"No of Iteration: ";
cin>>n;
f r=M_PI*log(2+sqrt(3))/8+3/8;
i=1;
f a=0.5,term=0.5,p=0.5;
while(i<=n){
p=p*i/(2*i+1);
term=term*i/(2*i+1);
term=term+p/(2*i+1);
a+=term;
i++;
}
cout<<"Catalan's constant for "<<n<<"th iteration: "<<setprecision(50)<<a<<endl;
}
return 0;
}
Code 3 (Best : Ramanujan's Formula Based Algorithm)
#include<bits/stdc++.h>
using namespace std;
typedef long long l;
typedef long double f;
int main(){
l t,i,n;
cout<<"No of testcases: ";
cin>>t;
while(t--){
cout<<"No of Iteration: ";
cin>>n;
f r=M_PI*log(2+sqrt(3))/8+3/8;
i=1;
f a=1,term=1,p;
while(i<=n){
term=term*(2*i-1)*2/i;
p=term*pow(2*i+1,2);
a+=1/p;
i++;
}
a=r+a*3/8;
cout<<"Catalan's constant for "<<n<<"th iteration: "<<setprecision(50)<<a<<endl;
}
return 0;
}
Thanks
Team IIT KGP Study Material
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