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Day 25: Running Time and Complexity
Day 25: Running Time and Complexity
+ 0 comments C++ Solution:
int count = 0; cin >> count; for(int i = 0; i < count; i++) { int num; cin >> num; bool isPrime = true; if(num > 1) { for(int j = 2; j <= sqrt(num); j++) { if(num % j == 0) { isPrime = false; break; } } } else isPrime = false; if(isPrime) cout << "Prime" << endl; else cout << "Not prime" << endl;
+ 0 comments My c++ solution.
bool isPrime(int n){ bool prime = true; if (n<2) prime = false; else if(n%2==0 && n!=2) prime = false; else { for(int i=3;i<n;i+=2){ if(n%i==0){ prime=false; break; } } } return prime; } int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int t,n; cin >> t; for(int i=0;i<t;i++){ cin>>n; if(isPrime(n)) cout<<"Prime"<<endl; else cout<<"Not prime"<<endl; } return 0; }
+ 0 comments import math def is_prime(n): if n <= 1: return "Not prime" if n == 2: return "Prime" if n % 2 == 0: return "Not prime" sqrt_n = int(math.sqrt(n)) for i in range(3, sqrt_n + 1, 2): if n % i == 0: return "Not prime" return "Prime" T = int(input()) for _ in range(T): n = int(input()) result = is_prime(n) print(result)
+ 0 comments C++
bool isPrime(int a){ bool prime = true; if(a < 2) prime = false; else if(a == 2) prime = true; else if(a % 2 == 0) prime = false; // If a is divisible by some number smaller than its square root, // then it is not prime else{ for(int i = 3; i <= sqrt(a); i += 2){ if(a % i == 0){ prime = false; break; } } } return prime; } int main(){ int a = 0; int t = 0; bool prime = true; cin >> t; while(t-- > 0){ cin >> a; prime = isPrime(a); if(prime) cout << "Prime" << endl; else cout << "Not prime" << endl; } return 0; }
+ 0 comments # Enter your code here. Read input from STDIN. Print output to STDOUT t=int(input()) for _ in range(t): i=int(input()) if i<=1: print('Not prime') elif i<=3: print('Prime') elif i%2==0 or i%3==0: print('Not prime') else: n=5 while n*n<=i: if i%n==0 or i%(n+2)==0: print('Not prime') break n+=6 else: print('Prime')
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