传送门

分析

首先涉及到环形问题的时候，我们常用的解决方法就是开两倍空间把这个环给储存下来

然后我们可以正着反正扫一遍，记录每一种颜色的珠子的最长前缀和最长后缀，最后再扫描一下答案即可

代码

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 1e3 + 10;
string str;
int n;
int bl[N];
int br[N];
int rl[N];
int rr[N];

int main(){
    scanf("%d",&n);
    cin >> str;
    str = str + str;
    // if(str[0] == 'b') bl[0] = 1;
    // if(str[0] == 'r') rl[0] = 1;
    // if(str[0] == 'w') rl[0] = 1,bl[0] = 1;
    for(int i = 0;i < n * 2;i++){
        if(str[i] == 'b')
            bl[i] = bl[i - 1] + 1,rl[i] = 0;
        if(str[i] == 'r')
            rl[i] = rl[i - 1] + 1,bl[i] = 0;
        if(str[i] == 'w')
            bl[i] = bl[i - 1] + 1,rl[i] = rl[i - 1] + 1;
    }
    for(int i = n * 2 - 1;i >= 0;i--){
        if(str[i] == 'b')
            br[i] = br[i + 1] + 1,rr[i] = 0;
        if(str[i] == 'r')
            rr[i] = rr[i + 1] + 1,br[i] = 0;
        if(str[i] == 'w')
            br[i] = br[i + 1] + 1,rr[i] = rr[i + 1] + 1;
    }
    int ans = 0;
    for(int i = 0;i < 2 * n;i++){
        int t = max(bl[i - 1],rl[i - 1]) + max(br[i],rr[i]);
        ans = max(t,ans);
    }
    printf("%d\n",min(ans,n));
}