A -
Hamburgers
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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 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.
Sample Input
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
简单二分。
这是我们周赛时用vj扒的题，当时比赛时思路就是各种分情况，想想有点晕就没做，后来提醒说是二分，才自己想明白了，挺简单。二分求上界。
#include <stdio.h>
#include <string.h>
#include <map>
#include <queue>
#include <algorithm>
using namespace std;
typedef __int64 ll;
const ll MAX=1000000000000+100;
ll need[4],pos[4],pri[4],money; //need做一个汉堡需要材料的量，pos拥有的，pri价格，money拥有的钱
char s[105];
//判断是否能够做出x个汉堡
bool C(ll x){
ll i,sum=0;
for(i=1;i<=3;i++)
if(pos[i]<x*need[i]) //如果拥有的材料不够做x个，则计算需要用多少钱来买该材料
sum+=pri[i]*(x*need[i]-pos[i]);
return money>=sum; //钱够返回true
}
ll upper_bound(){
ll l,r,mid;
l=0;r=MAX;
while(l<r){
mid=(l+r)/2+1; //加1处理可以防止死循环，--b
if(C(mid))
l=mid;
else
r=mid-1;
}
return l;
}
int main()
{
ll i,j,k;
scanf("%s",s);
scanf("%I64d%I64d%I64d",&pos[1],&pos[2],&pos[3]);
scanf("%I64d%I64d%I64d%I64d",&pri[1],&pri[2],&pri[3],&money);
memset(need,0,sizeof(need));
for(i=0;s[i]!='\0';i++){
if(s[i]=='B') need[1]++;
if(s[i]=='S') need[2]++;
if(s[i]=='C') need[3]++;
}
printf("%I64d\n",upper_bound());
return 0;
}