Codeforces Round #218 (Div. 2) C题

C. Hamburgers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.

Polycarpus has n_b pieces of bread, n_s pieces of sausage and n_c pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are p_b rubles for a piece of bread, p_s for a piece of sausage and p_c for a piece of cheese.

Polycarpus has r rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.

Input

The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).

The second line contains three integers n_b, n_s, n_c (1 ≤ n_b, n_s, n_c ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers p_b, p_s, p_c (1 ≤ p_b, p_s, p_c ≤ 100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer r (1 ≤ r ≤ 10¹²) — the number of rubles Polycarpus has.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.

Sample test(s)

Input

BBBSSC 6 4 1 1 2 3 4

Output

2

Input

BBC 1 10 1 1 10 1 21

Output

7

Input

BSC 1 1 1 1 1 3 1000000000000

Output

200000000001

题意：就是给告诉你一个产品需要3种材料，然后给了你做一个产品各种材料需要多少以及你现在拥有的材料数和钱，让你求出最多可以做出多少产品。
分析：比赛的时候根本就没往2分方面想，然后一直在分类讨论，导致结果越分越多，最后把自己搞晕了，其实在之前我做过类似的题目，我这次竟然没往2分这方面去想！！
代码实现：

#include<stdio.h>
#include<string.h>
#include<math.h>
char str[105];
__int64 n1,n2,n3,need1,need2,need3;
__int64 cost1,cost2,cost3,money,l,r;

void solve()
{
    __int64 mid,temp,res;
    l=0;r=money+n1+n2+n3;
    temp=money;
    while(l<=r)
    {
        mid=(l+r)/2;
        money=temp;
        if(mid*need1>n1)
         money=money-(mid*need1-n1)*cost1;
        if(mid*need2>n2)
         money=money-(mid*need2-n2)*cost2;
        if(mid*need3>n3)
         money=money-(mid*need3-n3)*cost3;
        if(money<0)
         r=mid-1;
        else
        {
            res=mid;
            l=mid+1;
        }
    }
    printf("%I64d\n",res);
}

int main()
{
    __int64 i;
    scanf("%s",str);
    need1=need2=need3=0;
    scanf("%I64d%I64d%I64d",&n1,&n2,&n3);
    scanf("%I64d%I64d%I64d%I64d",&cost1,&cost2,&cost3,&money);
    for(i=0;str[i]!='\0';i++)
     if(str[i]=='B')
      need1++;
     else if(str[i]=='S')
      need2++;
     else
      need3++;
    solve();
    return 0;
}