N-th Prime

Discussions Toph Problems
27

i’m getting WA on test 3
#include<bits/stdc++.h>
using namespace std;
const int N = 500000 ;
long long a[N];
int main(){
a[0] = 2;
a[1] = 3;
long long n,s,i,j=1;
for(n=4;n<=N;n++){
for(i=2;i<=sqrt(n);i++){
s = n%i;
if(s==0)
break;
}
if(s!=0){
j++;
a[j] = n;
}
}
long long k;
cin >> k;
cout << a[k-1] << endl;
return 0;
}

28

Need help to find out mistake

mxn=int(5e5)
vec=[0]*(mxn+1)

for i in range(2, len(vec)):
    if vec[i]: continue
    if i*i>mxn: break
    for j in range(i*i, len(vec), i):
        if vec[j]: continue
        vec[j]=1

n=int(input())
c=0
for i in range(2, len(vec)):
    if vec[i]: continue
    c+=1
    if c==n: break

print(i)
30

Here is my code. Why is it showing CPU Limit Exceeded? How can it be optimised?

#include<bits/stdc++.h>
using namespace std;

bool isprime(int x){
    bool k=true;
    for(int i=7;i*i<=x;i+=30){
        if(k%i==0 || k%(i+4)==0 || k%(i+6)==0 || k%(i+10)==0 || k%(i+12)==0 || k%(i+16)==0 || k%(i+22)==0 || k%(i+24)==0) {
            k=false; 
            break;
        }
    }
    return k;
}

int main (){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    long n;
    cin >> n;
    if(n<4){
        if(n==1) cout << 2;
        if(n==2) cout << 3;
        if(n==3) cout << 5;
    }

    long prime=0, s=7;
    n-=3;
    while(n!=0){
        if(isprime(s)){
            prime = s;
            n--;
        }
        if(n!=0 && isprime(s+4)){
            prime = s+4;
            n--;
        }
        if(n!=0 && isprime(s+6)){
            prime = s+6;
            n--;
        }
        if(n!=0 && isprime(s+10)){
            prime = s+10;
            n--;
        }
        if(n!=0 && isprime(s+12)){
            prime = s+12;
            n--;
        }
        if(n!=0 && isprime(s+16)){
            prime = s+16;
            n--;
        }
        if(n!=0 && isprime(s+22)){
            prime = s+22;
            n--;
        }
        if(n!=0 && isprime(s+24)){
            prime = s+24;
            n--;
        }
        s+=30;
    }

    cout << prime;
    return 0;

}
31 
from math import sqrt


def isPrime(x):
    if x == 2:
        return True
    if x < 2 or x % 2 == 0:
        return False
    i = 3
    while i < int(sqrt(x)):
        if x % i == 0:
            return False
        i += 2
    return True


def nThPrime(n):
    result=0
    if n<1: return result
    if n==1: return 2
    count = 1
    j = 3
    while count != n:
        if isPrime(j):
            count += 1
            result=j
        j += 2
    return result

n = int(input())
print(nThPrime(n))

I think, I am the person that you were talking about😞. can you help me?
How do I make it more efficient?

33

Can anyone tell me, how can I solve it in python?
I have done it in c with the “Sieve of Eratosthenes” but failed in python!