Convert PARI program to C++ -

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i found sequence of interest in oeis , want generate same sequence in c++ programming competition solution working on.

however hit roadblock understanding how program given in sequence page works.

here program given in page - 

(pari) test(n)= {m=n; forprime(p=2, 5, while(m%p==0, m=m/p));                                          return(m==1)} for(n=1, 500, if(test(n), print1(n", "))) (pari) a(n)=local(m); if(n<1, 0, n=a(n-1);              until(if(m=n, forprime(p=2, 5, while(m%p==0, m/=p)); m==1), n++); n) (pari) list(lim)={ lim\=1; my(v=list(), s, t); for(i=0, log(lim+.5)\log(5),     t=5^i;     for(j=0, log(lim\t+.5)\log(3),         s=t*3^j;         while(s <= lim,             listput(v, s);             s <<= 1;         )     ) ); vecsort(vec(v)) };

i found out pari is, unable convert program c++. suggestions me generate same sequence in c++ appreciated.

i tried generate sequence in c++ following code snippet. think missing numbers in between fail few tests in online ide.

for(int = 0; < 16; i++) {     for(int j = 0; j < 15; j++)     {         for(int k = 0; k < 12; k++)         {             std::cout<<pow(2,i)*pow(3,j)*pow(5,k)<<std::endl;         }     } }

i chose 16, 15 , 12 limits because otherwise result value overflows long variable type.

you have 3 programs here, each of serve different purposes.

the first checks if number 5-smooth. divides 2, 3, , 5 until can't more, , tests if what's left 1.

the second generates ''n''th 5-smooth number. uses same idea first, testing each number in range. inefficient!

the third generates 5-smooth numbers given bound.

i'm going assume third want, because seems applicable situation. (it helps author of program.)

#include <iostream> #include <vector> #include <algorithm>  int main(void); std::vector<long> smooth(long lim);  int main(void) {     long lim = 1000;     std::vector<long> v = smooth(lim);     std::cout << "5-smooth numbers " << lim << ": ";     (std::vector<long>::iterator = v.begin(); != v.end(); it++) {         std::cout << *it << ", ";     }     std::cout << "\n";     return 0; }  std::vector<long> smooth(long lim) {     std::vector<long> v = {};     (long t = 1; t <= lim; t*=5) {         (long s = t; s <= lim; s*=3) {             (long n = s; n <= lim; n*=2) {                 v.push_back(n);             }         }     }     std::sort(v.begin(), v.end());     return v; }

this isn't line-by-line conversion, of course; example, didn't use logarithms since exact logarithms aren't built-in c++ pari. it's pretty fast, finds 5-smooth numbers 1,844,674,407,370,955,161 (the highest can on 64-bit machine) in fraction of second.


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