Description
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 ≤ 10 12 ) — 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 Input
BBBSSC
6 4 1
1 2 3
4
2
BBC
1 10 1
1 10 1
21
7
BSC
1 1 1
1 1 3
1000000000000
200000000001
1
#include <stdio.h>
2
#include <
string
.h>
3
4
const
int
inf=
0x3f3f3f3f
;
5
const
long
long
INF=1e14+
7
;
6
char
a[
105
];
7
long
long
b,s,c,nb,ns,nc,pb,ps,pc,r;
8
9
bool
C(
long
long
x)
10
{
11
long
long
rb,rs,rc;
12
rb=(x*b-nb)>=
0
?(x*b-nb):
0
,rs=(x*s-ns)>=
0
?(x*s-ns):
0
,rc=(x*c-nc)>=
0
?(x*c-nc):
0
;
13
if
((rb*pb+rs*ps+rc*pc)<=
r)
14
return
true
;
15
else
16
return
false
;
17
}
18
19
int
main()
20
{
21
int
i,j,k;
22
while
(scanf(
"
%s
"
,a)!=
EOF)
23
{
24
b=
0
,s=
0
,c=
0
;
25
for
(i=
0
;a[i]!=
'
\0
'
;i++
)
26
{
27
if
(a[i]==
'
B
'
)
28
b++
;
29
else
if
(a[i]==
'
S
'
)
30
s++
;
31
else
if
(a[i]==
'
C
'
)
32
c++
;
33
}
34
//
printf("%d %d %d",b,s,c);
35
scanf(
"
%I64d %I64d %I64d %I64d %I64d %I64d
"
,&nb,&ns,&nc,&pb,&ps,&
pc);
36
scanf(
"
%I64d
"
,&
r);
37
long
long
lb=
0
,ub=
INF;
38
39
while
(ub-lb>
1
)
40
{
41
long
long
mid=(lb+ub)/
2
;
42
if
(C(mid))
43
lb=
mid;
44
else
45
ub=
mid;
46
}
47
48
printf(
"
%I64d\n
"
,lb);
49
}
50
51
return
0
;
52
}
