ZeptoLab Code Rush 2015---C. Om Nom and Candies

A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him to think a bit in order to enjoy candies the most. Would you succeed with the same task if you were on his place?

One day, when he came to his friend Evan, Om Nom didn’t find him at home but he found two bags with candies. The first was full of blue candies and the second bag was full of red candies. Om Nom knows that each red candy weighs Wr grams and each blue candy weighs Wb grams. Eating a single red candy gives Om Nom Hr joy units and eating a single blue candy gives Om Nom Hb joy units.

Candies are the most important thing in the world, but on the other hand overeating is not good. Om Nom knows if he eats more than C grams of candies, he will get sick. Om Nom thinks that it isn’t proper to leave candy leftovers, so he can only eat a whole candy. Om Nom is a great mathematician and he quickly determined how many candies of what type he should eat in order to get the maximum number of joy units. Can you repeat his achievement? You can assume that each bag contains more candies that Om Nom can eat.
Input

The single line contains five integers C, Hr, Hb, Wr, Wb (1 ≤ C, Hr, Hb, Wr, Wb ≤ 109).
Output

Print a single integer — the maximum number of joy units that Om Nom can get.
Sample test(s)
Input

10 3 5 2 3

Output

16

Note

In the sample test Om Nom can eat two candies of each type and thus get 16 joy units.

如果某种糖的W值 ≥C−−√， 那么可以直接枚举这种糖的数量

如果2种糖的W值均 ≤C−−√, 这时我们假设, Hb⋅Wr<Hr⋅Wb
设蓝色的糖拿了X个，且X≥Wr, 我们把X分为2份，X=Y+k⋅Wr, 由于 Hb⋅Wr<Hr⋅Wb，所以把k⋅Wr的那一部分拿来换成红色的糖，性价比更高，因此，在这种情况下，蓝色的糖的数目应该<Wr
Hb⋅Wr>Hr⋅Wb 和 Hb⋅Wr=Hr⋅Wb的情况同理可以推出

/*************************************************************************    > File Name: Z-C.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年04月05日 星期日 13时09分31秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace std;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;int main(){    LL C, Hr, Hb, Wr, Wb;    while (cin >> C >> Hr >> Hb >> Wr >> Wb)    {        if (C < Wr && C < Wb)        {            cout << 0 << endl;            continue;        }        if (C < Wr && C >= Wb)        {            cout << C / Wb * Hb << endl;            continue;        }        if (C < Wb && C >= Wr)        {            cout << C / Wr * Hr << endl;            continue;        }        LL ans = 0;        for (LL i = 0; i * i <= C; ++i)        {            if (C - i * Wb >= 0)            {                ans = max(ans, i * Hb + (C - i * Wb) / Wr * Hr);            }            if (C - i * Wr >= 0)            {                ans = max(ans, i * Hr + (C - i * Wr) / Wb * Hb);            }        }        cout << ans << endl;    }    return 0;}