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 : x y z

Line : the sequence consisting of only three characters 'R', 'Y' and 'B'. 

For % data, the length of the sequence is no more than .

For % data, the length of the sequence is no more than ,,  <= x, y, z <= .

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

RRYBRBRYBRY

样例输出

　　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 = , cy = , cb = ;
        int res = ;
        int tmp;
        for (int i = ;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 = ;
            }
        }
        tmp = cr + cy + cb;
        if(res < tmp)
            res = tmp;
        cout<<res<<endl;
    }
    return ;
}