BNUOJ-26476 Doorman 贪心

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=26476

　　题意：给一个字符序列，比如MWMMW，每次可以取前面两个中的一个，取出来后，取出来的那个个数加一，要求使得两个字符的个数不超过n，求最多能取多少个。。

　　贪心就可以了。

 //STATUS:C++_AC_36MS_1480KB

 #include <functional>

 #include <algorithm>

 #include <iostream>

 //#include <ext/rope>

 #include <fstream>

 #include <sstream>

 #include <iomanip>

 #include <numeric>

 #include <cstring>

 #include <cassert>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <bitset>

 #include <queue>

 #include <stack>

 #include <cmath>

 #include <ctime>

 #include <list>

 #include <set>

 //#include <map>

 using namespace std;

 //#pragma comment(linker,"/STACK:102400000,102400000")

 //using namespace __gnu_cxx;

 //define

 #define pii pair<int,int>

 #define mem(a,b) memset(a,b,sizeof(a))

 #define lson l,mid,rt<<1

 #define rson mid+1,r,rt<<1|1

 #define PI acos(-1.0)

 //typedef

 typedef long long LL;

 typedef unsigned long long ULL;

 //const

 const int N=;

 const int INF=0x3f3f3f3f;

 const int MOD=1e9+,STA=;

 //const LL LNF=1LL<<60;

 const double EPS=1e-;

 const double OO=1e15;

 const int dx[]={-,,,};

 const int dy[]={,,,-};

 const int day[]={,,,,,,,,,,,,};

 //Daily Use ...

 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 //End

 int n,vis[N],ans;

 char s[N];

 int main(){

  //   freopen("in.txt","r",stdin);

     int i,j,l,r;

     while(~scanf("%d",&n))

     {

         scanf("%s",s);

         mem(vis,);

         int len=strlen(s);

         int cntm,cntw;

         cntm=cntw=ans=;

         for(i=;i<len;i++){

             l=r=-;

             for(j=;j<len;j++){

                 if(!vis[j]){

                     if(l==-)l=j;

                     else {r=j;break;}

                 }

             }

             if(cntm<=cntw){

                 if(s[l]=='M')cntm++,vis[l]=;

                 else if(s[r]=='M')cntm++,vis[r]=;

                 else cntw++,vis[l]=;

             }

             else {

                 if(s[l]=='W')cntw++,vis[l]=;

                 else if(s[r]=='W')cntw++,vis[r]=;

                 else cntm++,vis[l]=;

             }

             if(abs(cntm-cntw)>n)break;

             ans++;

         }

         printf("%d\n",ans);

     }

     return ;

 }