CF371C（二分）

地址：http://codeforces.com/contest/371/problem/C

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 nb pieces of bread, ns pieces of sausage and nc pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are pb rubles for a piece of bread, ps for a piece of sausage and pc 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 nb, ns, nc (1 ≤ nb, ns, nc ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers pb, ps, pc (1 ≤ pb, ps, pc ≤ 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 %I64dspecifier.

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：做汉堡包，需要的材料用字符串给出了，已有的材料数在第二行，商店卖的材料的出钱在第三行，第四行是你有的钱数。问最多做多少汉堡。

思路：一开始自己做，做到一半发现要考虑的条件太多，思维不严谨的话会丢失一些可能性而错误，所以放弃了。赛后找了大神的代码看了下，27行，真简便o.0，这里把按大神思路写的代码写出。找到最大可能值，最小可能值，二分查找最终答案。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
//#include<algorithm>
using namespace std;
#define LL __int64
int main()
{
    char s[110];
    gets(s);
    LL hb=0,hs=0,hc=0;
    for(int i=0;i<strlen(s);i++)
    {
        if(s[i]=='B') hb++;
        if(s[i]=='S') hs++;
        if(s[i]=='C') hc++;
    }
    LL nb,ns,nc;scanf("%I64d%I64d%I64d",&nb,&ns,&nc);
    LL pb,ps,pc;scanf("%I64d%I64d%I64d",&pb,&ps,&pc);
    LL r;scanf("%I64d",&r);
    LL ll=0,rr=r+nb+ns+nc; //关键在最大值确定的问题上，最大的可能值是你有的钱数加上你有的材料数
    while(rr-ll>1)
    {
        LL mid=(ll+rr)/2,t=r;
        if(mid*hb>nb) t-=(mid*hb-nb)*pb;
        if(mid*hs>ns) t-=(mid*hs-ns)*ps;
        if(mid*hc>nc) t-=(mid*hc-nc)*pc;
        if(t>=0) ll=mid;else rr=mid;
    }printf("%I64d\n",ll);
    return 0;
}