去掉题目的背景：就是一个环形的串中，寻找一个最长的子串，该串由前后由俩部分组成，连续的b串和连续的r串，当然，一种颜色也可以；w可转变成任意颜色；

我的思路：比较简单的思路，但是是一个复杂度O（n^2）的算法；

先求出以每一个位置开始的最长串的长度，再找出可衔接的最长串即可；

USACO上面有更快的O（n)算法。

Holland's Frank Takes has a potentially easier solution: /* This solution simply changes the string s into ss, then for every starting// symbol it checks if it can make a sequence simply by repeatedly checking // if a sequence can be found that is longer than the current maximum one.*/#include 
     
      #include 
      
       using namespace std;int main() {  fstream input, output;  string inputFilename = "beads.in", outputFilename = "beads.out";    input.open(inputFilename.c_str(), ios::in);  output.open(outputFilename.c_str(), ios::out);    int n, max=0, current, state, i, j;  string s;  char c;    input >> n >> s;  s = s+s;  for(i=0; i
       
         max)      max = current;  } // for      output << max << endl;  return 0;} // main
       
      
     

/*ID: nanke691LANG: C++TASK: beads*/#include
     
      #include 
      
       #include
       
        using namespace std;char s1[700];int dp[355];int main(){	int n;    FILE *fin  = fopen ("beads.in", "r");    FILE *fout = fopen ("beads.out", "w");	//while(scanf("%d",&n)==1)	//{	    fscanf(fin,"%d",&n);		fscanf(fin,"%s",s1);		for(int i=0;i
        
         n) 				sum=dp[i]+dp[t-n];			else sum=dp[i]+dp[t+i];			if(sum>max1)				max1=sum;			if(max1>=n)			{				max1=n;				break;			}		}		fprintf(fout,"%d\n",max1);	//}	return 0;}