Codeforces 371C：Hamburgers（二分）

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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.

Examples
input
BBBSSC6 4 11 2 34
output
2
input
BBC1 10 11 10 121
output
7
input
BSC1 1 11 1 31000000000000
output
200000000001题目大意：给你一个汉堡从上到下的结构，由B、S、C三种食材组成，然后给出你他拥有多少个B、S、C的数量，再下一行给出每个B、S、C的单价，最后一行给出你他的钱数，问最多能做多少个这样的汉堡？
解题思路：统计B、S、C的个数，然后二分法，做mid块需要多少钱，拿需要的钱和已有的钱比较，来取舍区间。（注意某种结构中不含B、S、C的元素的情况）
代码如下：
#include <cstdio>#include <cstring>#define LL __int64LL nb,ns,nc,pb,ps,pc,r,numb=0,nums=0,numc=0;bool judge(LL x){if(numb==0)//不含b ，联系下列式子将pb临时变成0 pb=0;if(nums==0)//不含s ps=0;if(numc==0)//不含c pc=0;LL needmoneyb=(numb*x-nb)*pb;if(needmoneyb<0)//制作x个汉堡不再需要B 这种食材 。不写的话变负数，会影响二分区间的取舍 {needmoneyb=0;}LL needmoneys=(nums*x-ns)*ps;if(needmoneys<0){needmoneys=0;}LL needmoneyc=(numc*x-nc)*pc;if(needmoneyc<0){needmoneyc=0;}LL needmoney=needmoneys+needmoneyb+needmoneyc;if(r>=needmoney){return true;}else{return false;}}int main(){char a[110];scanf("%s",a);LL size=strlen(a);for(LL i=0;i<size;i++){if(a[i]=='B'){numb++;}if(a[i]=='S'){nums++;}if(a[i]=='C'){numc++;}}scanf("%I64d%I64d%I64d",&nb,&ns,&nc);scanf("%I64d%I64d%I64d",&pb,&ps,&pc);scanf("%I64d",&r);LL l=0,r=10000000000000;LL mid,ans;while(l<=r){mid=(l+r)/2;if(judge(mid)){ans=mid;l=mid+1;}else{r=mid-1;}}printf("%I64d\n",ans);return 0;}

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