鏈接：
Doorman
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.
Input
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.
Output
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.
Sample Input
1 MWWMWMMWM
Sample Output
9
Source
題意：
思路：
要一直判斷到男女之差 > N+1
因爲即便是最多隻能容許 N 個人沒有配對,查找到了第 N+1 個人不能配對
那麼如果查到下一個人可以和前面的配對,那麼就可以新增加兩個人
也就是說：
2
MMMW 的結果是 4
code：
/**
題意：第一行 N 表示能夠容忍的不能配對的人數
第二行一條 100 個字符以內的字符串'W'代表女人, 'M'代表男人
看門人一次只讓一個人進去,男女儘量配對,如果不配對的人數超過了 N
那麼看們人就分不清了,此時他就會宣佈俱樂部人滿,不再讓後面的人進入
輸出能進去的最多的人數。
注意：關於分不清的時候的跳出判斷情況。
要一直判斷到男女之差 > N+1
因爲即便是最多隻能容許 N 個人沒有配對,查找到了第 N+1 個人不能配對
那麼如果查到下一個人可以和前面的配對,那麼就可以新增加兩個人
也就是說：
2
MMMW 的結果是 4
*/
#include<string>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[110];
int n;
int main()
{
while(scanf("%d", &n) != EOF)
{
scanf("%s", &str);
int n1;
int n2;
int ans = 0;
n1 = n2 = 0;
int len = strlen(str);
int flag = 0;
int Min, Max;
for(int i = 0; i < len; i++)
{
if(str[i] == 'M') n1++;
if(str[i] == 'W') n2++;
Min = min(n1, n2); //依次記錄一次
Max = max(n1, n2);
if(Max-Min > (n+1)) //如果男女之差比能准許的 N 要大 2,那麼守門人就分不清了,馬上宣佈人滿,不管後面排隊的
{
flag = 1; //標記
break;
}
}
if(flag == 0 && Max-Min <= n) //如果沒有標記,全部排隊的人數也合法：男女之差不超過 N
ans = len;
else //如果標記了,最大限度的讓人進去,男女之差正好爲 N
ans = Min*2+n; //當然也包含了沒有標記,男女之差 > N 的情況
printf("%d\n", ans);
}
return 0;
}