程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> CF:371C - Hamburgers 二分答案暴力解決

CF:371C - Hamburgers 二分答案暴力解決

編輯:C++入門知識

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

思路:直接枚舉答案,然後計算出他們的價值,如果大於給的錢上界就等於mid,否則下界等於mid,如此便可找出最佳答案……

這樣的題可以二分,以前竟然不知道,唉……確實如魏神所說,做題還是太少了呀!!!以前也沒有用二分做這些題,再在終於知道二分的用處有多大了,只要類似這種題都可以直接用二分暴力找出最佳答案的。

不過在上下界的判斷語句上還不是很熟……前閉後開,前閉後閉的二分還不是分得很清楚,所以這題while中是 l+1

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define PI acos(-1.0)
#define eps 1e13
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 100005
#define MN 1005
#define INF 10000007
using namespace std;
typedef long long ll;
ll ss[4],z[4],zz[4];
int main()
{
    char s[102];
    ll i,p,l=0,r=eps,mid,cost;
    scanf("%s",s);
    for(i=0;i>z[0]>>z[1]>>z[2]>>zz[0]>>zz[1]>>zz[2]>>p;
    while(l+1>1;
        cost=0;
        for(i=0;i<3;i++)
            if(mid*ss[i]>z[i]) cost+=(mid*ss[i]-z[i])*zz[i];
        if(cost<=p) l=mid;
        else r=mid;
    }
    cout<

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved