問題 E: Doorman

時間限制: 1 Sec  記憶體限制: 128 MB 提交: 51  解決: 23 [提交][狀態][討論版][命題人:外部匯入]

題目描述

The doorman Bruno at the popular night club Heaven is having a hard time fulfilling his duties. He was told by the owner that when the club is full, the number of women and men let into the club should be roughly the same. When the night club opens, people wanting to enter the club are already lined up in a queue, and Bruno can only let them in one-by-one. He lets them in more-or-less in the order they are lined up. He can however decide to let the second person in the queue cut the line and slip into the club before the person in front. This will no doubt upset the person first in line, especially when this happens multiple times,but Bruno is quite a big guy and is capable of handling troublemakers. Unfortunately though, he is not that strong on mental calculations under these circumstances.He finds keeping track of the difference of the number of women and number of men let into the club a challenging task. As soon as the absolute difference gets too big, he looses track of his counting and must declare to the party people remaining in the queue that the club is full.

輸入

The first line of input contains a positive integer X< 100 describing the largest absolute difference between the number of women and number of men let into the club, that Bruno can handle. The second line contains a string consisting solely of the characters ’W’ and ’M’ of length at most 100, describing the genders of the people in the queue, in order. The leftmost character of the string is the gender of the person first in line.

輸出

The maximum number of people Bruno can let into the club without loosing track of his counting. You may assume that the club is large enough to hold all the people in the queue.

樣例輸入

2
WMMMMWWMMMWWMW

樣例輸出

8

/*有點做閱讀理解的意思，題目大體意思就是一個俱樂部門口有一些人在排隊，
門衛得保證進去的男女人數差 ，當不配對的大於x+1時，就關閉.
注意這裡是X+1，因為當前為x時，可能來的下一個人性別剛好與之前的性別配對，
那麼他們兩個就可以進去俱樂部 
*/
#include<bits/stdc++.h>
using namespace std;
char str[1000];
int main()
{
	int x;
	int w=0,m=0;
	int mins=-1,flag=0;
	cin>>x;
	getchar();
	cin>>str;
	int len=strlen(str);
	for(int i=0;i<len;i++)
	{
		if(str[i]=='W')
		w++;//當前women人數 
		else if(str[i]=='M')
		m++;//當前men人數 
		mins=min(w,m);//配對 
		if(fabs(w-m)>x+1)//不配對人數大於x+1，關閉俱樂部 
		{
			flag=1;
			break;
		}
	}
	if(flag)
	cout<<mins*2+x<<endl;
	else
	cout<<len<<endl;
return 0;	
}