zoukankan      html  css  js  c++  java
  • 微软2016校园招聘在线笔试之Magic Box

    题目1 : Magic Box
    时间限制:10000ms
    
    单点时限:1000ms
    
    内存限制:256MB

    描述

    The circus clown Sunny has a magic box. When the circus is performing, Sunny puts some balls into the box one by one. The balls are in three colors: red(R), yellow(Y) and blue(B). Let Cr, Cy, Cb denote the numbers of red, yellow, blue balls in the box. Whenever the differences among Cr, Cy, Cb happen to be x, y, z, all balls in the box vanish. Given x, y, z and the sequence in which Sunny put the balls, you are to find what is the maximum number of balls in the box ever.

    For example, let's assume x=1, y=2, z=3 and the sequence is RRYBRBRYBRY. After Sunny puts the first 7 balls, RRYBRBR, into the box, Cr, Cy, Cb are 4, 1, 2 respectively. The differences are exactly 1, 2, 3. (|Cr-Cy|=3, |Cy-Cb|=1, |Cb-Cr|=2) Then all the 7 balls vanish. Finally there are 4 balls in the box, after Sunny puts the remaining balls. So the box contains 7 balls at most, after Sunny puts the first 7 balls and before they vanish.

    输入

    Line 1: x y z
    
    Line 2: the sequence consisting of only three characters 'R', 'Y' and 'B'. 
    
    For 30% data, the length of the sequence is no more than 200.
    
    For 100% data, the length of the sequence is no more than 20,000, 0 <= x, y, z <= 20.

    输出
    The maximum number of balls in the box ever.

    提示

    Another Sample

    Sample Input Sample Output
    0 0 0 RBYRRBY             4

     样例输入

    1 2 3
    RRYBRBRYBRY

    样例输出

    7

      第一次使用hihoCoder,不了解怎么使用,吃了大亏。贴一下我事后写的代码,未经验证。

    solution:

    #include <iostream>
    #include <string>
    using namespace std;
    int main()
    {
        int x,y,z;
        string s;
        while (cin>>x>>y>>z>>s)
        {
            int cr = 0, cy = 0, cb = 0;
            int res = 0;
            int tmp;
            for (int i = 0;i < s.size();++i)
            {
                if(s[i] == 'R')
                    cr++;
                else if(s[i] == 'Y')
                    cy++;
                else
                    cb++;
                if(abs(cr-cy) != z)
                    continue;
                if(abs(cy-cb) != x)
                    continue;
                if(abs(cb-cr) != y)
                    continue;
                tmp = cr + cy + cb;
                if(res < tmp)
                {
                    res = tmp;
                    cr = cy = cb = 0;
                }
            }
            tmp = cr + cy + cb;
            if(res < tmp)
                res = tmp;
            cout<<res<<endl;
        }
        return 0;
    }
  • 相关阅读:
    英语面试-Behavioral Question
    算法与数据结构-各种循环的时间复杂度分析(英语)
    英语面试-Behavioral Question
    CS硕士全日制考研资料(含完整复习计划)
    算法与数据结构-树-简单-合并二叉树
    算法与数据结构-树-简单-二叉树的所有路径
    gcd的理解
    数学总结
    图论总结
    数据结构总结
  • 原文地址:https://www.cnblogs.com/gattaca/p/4391153.html
Copyright © 2011-2022 走看看