Codeforces Round#218(Div.2) C. Hamburgers

C. Hamburgers

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 %I64d specifier.

Output

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

Examples

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

题解： 给出汉堡的字符串（即是所需要的原料的种类）用给出nb,ns,nc 已有的材料，以及每中原料的价格pb,ps,pc 以及现有的钱数，求最多可以做多少汉堡

所以可以用二分法去求解，不断求最大的然后判断，假设可做n个， 看看每种原料的钱的总和够不够总钱数，当本身已有的材料满足条件时，那么该种类的钱设置为0。
AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#define LL long long#define inf 2000000000000using namespace std;char str[100];LL nb,ns,nc;LL num,pb,ps,pc;LL b=0,s=0,c=0;int judge(LL n){    LL bb=(n*b-nb)*pb,ss=(n*s-ns)*ps,cc=(n*c-nc)*pc;    if(bb<0) bb=0;    if(ss<0) ss=0;    if(cc<0) cc=0;    if(bb+ss+cc<=num) return 1;    else return 0;}int main(){    scanf("%s",str);    scanf("%lld%lld%lld",&nb,&ns,&nc);    scanf("%lld%lld%lld",&pb,&ps,&pc);    scanf("%I64d",&num);    for(int i=0;i<strlen(str);i++)    {        if(str[i]=='B') b++;        else if(str[i]=='S') s++;        else if(str[i]=='C') c++;    }    LL l=1,r=inf;    while(l<r)    {        LL mid=(l+r)>>1;        if(judge(mid))        {            l=mid+1;        }        else        {            r=mid;        }    }    printf("%I64d\n",l-1);    return 0;}