OK, I have been out of the loop with C++ for a year now. I am finished with school, so this is NOT homework. Just trying to clean up some old assignments for future references.
(the code is in "messy" shape til I can get it working properly and clean it up. that's why there are so many cout statements for checking what the hell is going on).
Running the program you see it asks for number of voters, and their names, and how many votes were received. That parts all good. When I do

perc = votes[i] /2;

or

perc = votes[i] * 100;

it works properly. But the

perc = (votes[i] / sum) *100;

to get the percentage, it's failing out. Like I said, probably an easy fix, just not seeing it. Again, out of the loop. Had java II, PHP, and ASP.NET since then and it's killing me that this is so simple and I'm not seeing it. So any advice is GREATLY appreciated.
the code:

//This program Dynamicly allocates memory for user to input number of array members for input
//data of an election and outputs names, number of votes, percentage, total, and winner.
//
//10-19-2010

#include <iostream>
#include <string>
using namespace std;

int main()
{ 
	
	string *name, // pointer to name array
		winner;
      int *votes; // pointer to vote array
	 
	int	i,
		largest, 
		sum = 0,
		*perc,
		total= 0,     
	    numVotes; 


	cout << "How many voters are there?: ";                    
	cin >> numVotes;
	cout<<endl;
    cin.ignore();
    name = new string[numVotes];  // Dynamicly allocate names
    
	for(i=0;i < numVotes; i++){
		cout<<"Enter canidates last names: ";
		getline(cin,name[i]);
	}
	perc = new int[numVotes];      // Dynamicly allocated percents
	votes = new int[numVotes];     // Dynamicly allocate votes
    cout<<endl;

	for(i = 0;i < numVotes; i++){                              // loop for biggest vote                 
		cout << "Enter votes received for "<< name[i] << ": ";             
		cin >> votes[i];
		total = total + votes[i];
	}    
     largest = votes[0];
	 winner = name[0];
	 for(i=0;i<numVotes;i++){
		 if(votes[i]>largest){
			 largest = votes[i];
			 winner = name[i];
		 }
	 }                                                      //end loop for biggest vote/winner
		
    cout<<endl;
    cout << "Canidate\t" << "Votes Received " << "\t% of Total Votes"<<endl;

	for(i=0;i < numVotes; i++){

        total = votes[i];	// get total
		sum = sum + total;
//perc[i] = (votes[i]/ sum)* 100;
		//cout << "percent is: " << perc[i]  << " : voting numbers are " << votes[i] << " " << " : the sum is " << sum << endl;

		 // percent of votes entered by user (by total, number of votes, and used 100%)
		/*cout << fixed << showpoint;
		cout << name[i] << "   \t   " << votes[i] << "   \t   " << setprecision(2) << perc <<  endl;*/
	}   
/*------------------------------------AREA OF CONCERN------------------------------------------------------------------*/
	for(i=0;i < numVotes; i++){
		cout <<"sums are: " << " " << sum;
		perc[i] = (votes[i] / sum) * 100; // <-----Getting "0" as perc. tried just varaible as in "perc" as int and same output.
		cout << "percent is: " << perc[i]  << " : voting numbers are " << votes[i] << " " << " : the sum is " << sum << endl;
	}
/*--------------------------------------END CONCERN---------------------------------------------------------------------*/
cout<<"\nTotal:  \t\t" << sum <<endl;	
    cout <<"\nThe winner of the Election is " << winner << "."<< endl;

delete [] votes;	 // delete vote array out of memory
delete [] name;      // delete name array out of memory
delete [] perc;

cin.sync();
cin.peek();


}

I'm going by your comment rather than the vague "it's failing out":

>perc = (votes / sum) * 100;
So votes and sum are both integers. Keeping in mind that any precision will be truncated, you probably wanted this instead:

perc[i] = (votes[i] / (double)sum) * 100;

By converting either operand of the division to a floating-point type, you preserve the precision long enough for the multiplication to produce a non-zero result.

Thank you very much Narue. Tried to type cast it last night, just went about it the wrong way. 'preciate mate !

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