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 ;
 }