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 <iostream>
#include <fstream>
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<n; i++) {c = (char) s[i];if(c == 'w')state = 0;elsestate = 1;j = i;current = 0;while(state <= 2) { // dont go further in second string than starting position in first stringwhile(j<n+i && (s[j] == c || s[j] == 'w')) { current++;j++;} // whilestate++;c = s[j];} // whileif(current > max)max = current;} // foroutput << max << endl;return 0;
} // main

/*
ID: nanke691
LANG: C++
TASK: beads
*/
#include<iostream>
#include <fstream>
#include<string.h>
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;i++)s1[i+n]=s1[i];s1[2*n]='\0';//puts(s1);int flag=0;memset(dp,0,sizeof(dp));for(int i=0;i<n;i++){   if(s1[i]=='w')flag=1;else flag=0;char temp=s1[i];for(int j=i+1;j<2*n;j++){if(flag&&s1[j]!='w'){flag=0;temp=s1[j];}if(temp==s1[j]||s1[j]=='w')dp[i]++;else break;}dp[i]++;}//for(int i=0;i<n;i++)//   printf("%d ",dp[i]);//cout<<endl;int max1=-1;for(int i=0;i<n;i++){int t=dp[i],sum;if(t>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;
}