C++ using iomanip

I have the following program in C++ and I need to display the cents part of the amount only. I cannot seem to get it to work using iomanip.

#include <iostream>
#include <iomanip>

using namespace std;

int main ()
      cout << setiosflags(ios::fixed|ios::showpoint|ios::right);
      int quarters, dimes, nickels, pennies;
      float amount;

      cout << "How many quarters? " << endl;
      cin >> quarters;

      cout << "How many dimes? " << endl;
      cin >> dimes;

      cout << "How many nickels? " << endl;
      cin >> nickels;

      cout << "How many pennies? " << endl;
      cin >> pennies;

      amount = (quarters*.25) + (dimes*.10) + (nickels*.05) + (pennies*.01) ;

      cout << "You have " << setprecision(2) << amount << " much money " << endl;

      cout << "You have " << noshowpoint << setprecision(0) << amount << " dollars" << endl;

      cout << "You have " << noshowpoint << setprecision(2) << amount << " cents" << endl;

      return 0;
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sarabandeConnect With a Mentor Commented:
to add to above comments:

it probably is easiest to have two integer variables dollars and cents and do the rounding as Zoppo has shown by cast from double to int.:

int dollars = (int)amount;
int cents = (int)((amount - dollars+0.005)*100);
std::cout << "You have " << dollars << " dollars" << std::endl;
std::cout << "You have " << cents<< " cents" << std::endl;

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ZoppoConnect With a Mentor Commented:

I don't think it's possible to format the output using <iomanip> on cout, for floating point values IMO the number before the decimal point can't be completely supressed. So I think the only possibility you have is to calculate the value for cents, i.e. like this:
 cout << "You have " << (int)( 100.0 * ( 0.005 + amount ) ) % 100 << " cents" << endl;

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Here I added 0.005 to the calculated value to avoid rounging errors (i.e. without this the resulting int for 1.52 would be 51).

Hope this helps,

std::ios_base::precision -
•Using the default floating-point notation, the precision field specifies the maximum number of meaningful digits to display in total counting both those before and those after the decimal point. Notice that it is not a minimum, and therefore it does not pad the displayed number with trailing zeros if the number can be displayed with less digits than the precision.
•In both the fixed and scientific notations, the precision field specifies exactly how many digits to display after the decimal point, even if this includes trailing decimal zeros. The digits before the decimal point are not relevant for the precision in this case.

You may wish to review modf:
In their example, their output is: 3.141593 = 3.000000 + 0.141593
You know how to handle the whole number. For the decimal part, you would use a method similar to zoppos approach.
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