/* Ugly Numbers A number is considered to be ugly if its only prime factors are 2, 3 or 5. [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …] is the sequence of ugly numbers. Task: Create a program which takes a number n as input and checks if it's an ugly number. Hint: An ugly number U can be expressed as: U = 2^a * 3^b * 5^c, where a, b and c are non-negative integers. Bonus: Ptint all the ugly numbers in a given range. -----------------Important------------------- Enter the lower and the upper bound separated by a space for a range. example: 1 20 If you just want to check for a number just input your number. example: 12 --------------------------------------------- */ #include #include #include using namespace std; int has_space(string s){ int l(s.length()); for(int i=0;i!=l;i++){ int ch=(int)s.at(i); if(!(ch-(int)' ')) return 1; } return 0; } int space(string s){ int l(s.length()); for(int i(0);i!=l;i++){ int ch=(int)s.at(i); if(!(ch-(int)' ')) return i; } } int *range(string s){ int sp=space(s); int deb=stoi(s.substr(0,sp)); int fin=stoi(s.substr(sp+1)); int *arr = new int[3]; *(arr)=deb;*(arr+1)=fin; return arr; } int prime(int i){ if(i<=1){return 0;} if(i==2){return 1;} for(int t=2;t!=i;t++){ if(i%t==0) return 0; } return 1; } int ugly(int x){ if(x==1){return 1;} if(prime(x) && x>5){return 0;} for(int i=2;i!=x;i++){ if(prime(i) && !(x%i) && i>5){ return 0; } } return 1; } int main(){ string s; getline(cin,s); if(has_space(s)){ vector v; int *ra; ra=range(s); int r=ra[0]; for(;r!=ra[1]+1;r++){ if(ugly(r)) v.push_back(r); } if(v.empty()){ cout<<"There are no ugly numbers from "<::iterator i; for(i=v.begin();i!=v.end();i++) cout<<*i<<" "; cout<<"."; } delete []ra; }else{ int res=stoi(s); cout<<"You entered "<

ugly.cpp