• CF 371C


    http://codeforces.com/problemset/problem/371/C

                                                            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 nbnsnc (1 ≤ nb, ns, nc ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers pbpspc (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 cincout 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

    【题意】  给出做一个汉堡需要的 三种材料的数量   手里已有的每种的数量 和钱数  问最多能做多少汉堡

     一开始做的时候 是想把情况分成很多种算  结果发现分起来会很混乱

    其实只要用二分就很容易了

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    __int64 yiyou[4],p[4],xuyao[10],money;
    char s[102];
    
    int ok(long long  x)
    {
        __int64 sum=0;
        for(int i=1;i<=3;i++)
            {
                __int64  t=(x*xuyao[i]-yiyou[i])*p[i];
                if(t>0)
                   sum+=t;
            }
        if(money>=sum)
            return 1;
        return 0;
    
    }
    
    __int64 solve()
    {
        __int64 l=0,r=money+yiyou[1]+yiyou[2]+yiyou[3];
        __int64 best=0;
        while(l<=r)
        {
            __int64 mid=(l+r)/2;
    
            if(ok(mid))
            {
               l=mid+1;
               if(best<mid)
                   best=mid;
            }
            else
                r=mid-1;
        }
        return best;
    }
    
    int main()
    {
        int i,j,n,m,t;
        while(~scanf("%s",s))
        {
            int len =strlen(s);
            for(i=1;i<=3;i++)
                scanf("%I64d",&yiyou[i]);
            for(i=1;i<=3;i++)
                scanf("%I64d",&p[i]);
            scanf("%I64d",&money);
            memset(xuyao,0,sizeof(xuyao));
            for(i=0;i<len;i++)
            {
                if(s[i]=='B')
                    xuyao[1]++;
                if(s[i]=='S')
                    xuyao[2]++;
                if(s[i]=='C')
                    xuyao[3]++;
            }
            printf("%I64d
    ",solve());
        }
        return 0;
    }
  • 相关阅读:
    CentOS如何查看端口是被哪个应用/进程占用
    VS2010自带的性能分析工具分析.NET程序的性能
    Centos7安装RocketMQ及配置测试
    tbnet编译
    jquery动态生成html代码 怎么 获取id 或 class
    jQuery1.9+中删除了live以后的替代方法
    自定义jQuery插件Step by Step
    jQuery插件开发的两种方法及$.fn.extend的详解
    Informix 配置选项
    Informix ConnetionString Pool Size
  • 原文地址:https://www.cnblogs.com/assult/p/3485381.html
Copyright © 2020-2023  润新知