ACM - UVA 562 - dividing coins

Problem: 562 - Dividing coins
Solution: C++
Hints: http://www.algorithmist.com/index.php/UVa_562

#include<iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cctype>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <sstream>
#include <utility>
#include <set>
#include <math.h>
using namespace std; 
#define N 101 //maximum number of coins
#define M 25001 //maximum share of each person, max coin is 500c and max number of coins is 100

bool  v[M]; 
int iabs(int n)
{
    return (n > 0)? n : -n;
}
int main()
{
    int i, j, NumberOfProblems, NoOfCoins, m, sum, a[N];
    //cin number of problems
    cin >> NumberOfProblems;

    while(NumberOfProblems-- > 0)
    {
        //cin number of coins 
        cin >> NoOfCoins; 

        //cin the coins and calculate their sum
        for(sum = 0, i = 1; i <= NoOfCoins; i++)
            cin >> a[i], sum += a[i];

       //the share of each person (minus one if sum is odd)
        m = sum / 2;
        memset(v, 0, sizeof(v));
        v[0] = true;

        //fill the v table
        //for each coins of the coins check with true the elements off the table v that
        // can be reached by adding this coin//example adding first coin 3
        //v[0] = v[3] = true//adding 2
        // v[0] = v[2] = v[3] =v[5]
        for(i = 1; i <= NoOfCoins; i++)
            for(j = m; j >= 1; j--)
                //if element j in the v table is false see if it could be reached or no//if its true leave it
                if(!v[j])
                    v[j] = (a[i] <= j)? v[j-a[i]] :false; 

         //see which is the last element that can be reached within the rangeof the share of each person
        for(j = m; j >= 1; j--)
            if(v[j])
                break;

        //the rest is the difference between total money and total shares
        cout << iabs(sum - j * 2)<<endl;
    }
    return 0;
}