URI Solution 1262 Multiple Reading in C++ language

URI Online Judge | 1262

Multiple Reading

By TopCoder*  USA

Timelimit: 1

In many computer systems, multiple processes can read from the same resource during the same clock cycle, but only a single process can write to the resource during a clock cycle. Reads and writes cannot be mixed during the same clock cycle. Given a history of the reads and writes that occurred during a particular computation as a String trace, and an int procs representing the number of processes used by the computation, calculate the minimum duration of the computation in clock cycles. The trace represents each read as an 'R' and each write as a 'W'.

For example, if trace is "RWWRRR" and procs is 3, then the minimum number of clock cycles is 4: one for the first read, one each for the two writes, and one for the last group of reads.

Input

The input contains several test cases. Each test case is composed by two lines. The first line has a string made out of 1 to 50 characters, where each can be either 'R' or 'W'. The second line contains an integer P (1 ≤ P ≤ 10), which represents the number of processes as a direct indicator of how many read operations can be performed simultaneously. The input stream ends in EOF.

Output

For each test case determine and print the minimum number of clock cycles required to run the given trace. For further reference see the examples below.

Sample Input	Sample Output
RWWRRR 3 RWWRRRR 3 WWWWW 5 RRRRRRRRRR 4 RWRRWWRWRWRRRWWRRRRWRRWRRWRRRRRRRRRWRWRWRRRRWRRRRR 4	4 5 5 3 30

* This problem is of autorship and property of TopCoder (www.topcoder.com/tc) and adapted by bitfreeze for authorized use in URI OJ.
* Unauthorized reproduction of this problem statement without the prior written consent of TopCoder, Inc. is strictly prohibited.

URI Solution 1262 Multiple Reading in C++ language

#include <iostream>

#include <cstring>

using namespace std;

int main()

{

    string s;

    int p, size, ciclos, count;

    while(cin >> s >> p)

    {

        size = s.size();

        ciclos = 0; count = 0;

        for (int i = 0; i < size; ++i)

        {

            if(s[i] == 'R'){

                if(count == 0)

                    ciclos++;

                count++;

                if(count == p)

                    count = 0;

            }else{

                ciclos++;

                count = 0;

            }

        }

        printf("%i\n", ciclos);

    }

    return 0;

}

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