MKUHAR - Most Servings Meal

Problem Link

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

struct node
{
int x, y, pm, sm, pv, sv;
node(){}
node(int x, int y, int pm, int sm, int pv, int sv)
{
this->x = x;
this->y = y;
this->pm = pm;
this->sm = sm;
this->pv = pv;
this->sv = sv;
}
};

int n;
node arr[101];

/*
Find out if it is possible to have given number of
servings for each ingredient with given money
*/
bool isServingPossible(int totalServings, int money)
{
int required = 0;

/* check if we can have totalServings of each ingredient */
for(int i = 0; i < n; i++)
{
required = totalServings * arr[i].x;
required -= arr[i].y;
int minMoneySpent = 99999999;

int limit = ceil(required / (double)arr[i].sm);

for(int numSmallItem = 0; numSmallItem <= limit; numSmallItem++)
{
int totalSmallItem = numSmallItem * arr[i].sm;

int numLargeItem = (required > totalSmallItem)? ceil((required - totalSmallItem) / (double)arr[i].sv) : 0;

minMoneySpent = min(minMoneySpent, numSmallItem * arr[i].pm + numLargeItem * arr[i].pv);
}

money -= minMoneySpent;

/* if no money left return false */
if(money < 0)
{
return false;
}
}

return true;
}

/*
find out if mid number of servings possible with given money
if yes search in mid to e
otherwise search in e to mid - 1
*/

void solve(int s, int e, int money)
{
while(s < e)
{
int mid = s + (e - s + 1) / 2;

if(isServingPossible(mid, money))
{
s = mid;
}
else
{
e = mid - 1;
}
}

cout << s << endl;
}

int main()
{
int m, x, y, pm, sm, pv, sv;
cin >> n >> m;

for(int i = 0; i < n; i++)
{
cin >> x >> y >> sm >> pm >> sv >> pv;
arr[i] = node(x, y, pm, sm, pv, sv);
}

solve(0, m, m);

return 0;
}