Boost, converting between arbitrary precision floating point and integral types

I am utterly baffled as to how one should actually accomplish this, I have mostly managed to get my code to work until I try to run ceil() on a cpp_dec_float<0> type at which point it will explode on me and refuse to cooperate, after careful examination of boost documentation and google searching I have concluded that I cannot locate the answer on my own, thus I turn here to ask for help from infinitely more competent programmers.

#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/math/special_functions/pow.hpp>
#include <boost/chrono/ceil.hpp>
#include <boost/multiprecision/number.hpp>

namespace mp = boost::multiprecision;
using namespace std;
using namespace boost::math;
using namespace boost::chrono;
using Int = mp::cpp_int;
Int reward(Int baseReward, int percent);

int main(int argc, char** argv)
{

    reward = calcReward(immBase, percent);
    return 0;
}

Int calcReward(Int baseReward, int percent)
{
    using Dec  = mp::number<mp::cpp_dec_float<0>>;
    Dec mult   = 0.1+0.01*percent;
    Dec five   = 5;
    Dec base   = baseReward.convert_to<Dec>();
    Int result = ceil(five*base*mult);

    return result;
}

that's the code when stripped from unnecessary junk, the five variable is a dirty hack to get it to accept the multiplication with the arbitrary precision floating point value (as it just decided to say that there's no matching operator * otherwise). This is rather bemusing and quite frustrating, ceil().convert_to() didn't work either and spat out an arbitrary undecipherable error. (template errors are unreadable)

this code does require the -std=c++11 flag when compiling, as is probably obvious.