Hint 2: The The total number of minutes elapsed from hrs_in:min_in to a later time point on the same day hrs_out:min_out is: ( hrs_out * 60 + min_out ) - ( hrs_in * 60 + min_in )
cout<<"\nEnter Log-in for "<<days[5]<<": ";
cin>>hrs_in>>colon>>min_in;
cout<<"Enter Log-out for "<<days[5]<<": ";
cin>>hrs_out>>colon>>min_out;
// at this point, hrs_in:min_in and hrs_out:min_out
// contain the in and out times for Friday
int totalwork;
totalwork=(hrs_out*.60+min_out)-(hrs_in*.60+min_in); // * 0.60 ???
// (hrs_out*60+min_out)-(hrs_in*60+min_in) would give
// the total number of *minutes* of work on that one day.
Firstly, please always use code tags - select your code, then press the <> button on the right. It is much easier because it formats the code correctly, and we can quote line numbers.
You should use the double type for your variables if you are going to have multiplication / division by fractions. With division, always check for divide by 0.
With this part (better with line numbers here):
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cout<<"\n\n\nEnter Log-in for "<<days[0]<<": ";
cin>>hrs_in>>colon>>min_in;
cout<<"Enter Log-out for "<<days[0]<<": ";
cin>>hrs_out>>colon>>min_out;
cout<<"\nEnter Log-in for "<<days[1]<<": ";
cin>>hrs_in>>colon>>min_in;
cout<<"Enter Log-out for "<<days[1]<<": ";
cin>>hrs_out>>colon>>min_out;
Consider using a for loop, instead of the repeated code.
To make sure you always convert to an integer number of hours, try the following:
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int temp;
temp = mins % 60; //gives us the remainder when (mins / 60)
hours = (mins - temp) / 60; //hours will be an integer, and temp the remaining minutes