【CF1025C】Plasticine zebra（模拟）

题意：

n<=1e5

思路：可以证明答案即为极长交替出现的串长度之和

需要注意其实这个串是一个环，复制后再做

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   2100000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)
 #define oo 110000000000000

 char a[N],b[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void swap(int &x,int &y)
 {
     int t=x;x=y;y=t;
 }

 ull gcd(ull a, ull b)
 {
   return a ==  ? b : gcd(b % a, a);
 } 

 int main()
 {
 //    freopen("1.in","r",stdin);
 //    freopen("1.out","w",stdout);
     scanf("%s",a);
     int n=strlen(a);
     for(int i=;i<=n;i++) b[i]=a[i-];
     for(int i=n+;i<=n*-;i++) b[i]=b[i-n];
     int m=n*-;
     int t=;
     int w=;
     int ans=;
     while(t<n)
     {
         t++;
         if(w<t) w=t;
         int tmp=;
         while(w+<=m&&w+<=t+n-&&b[w]!=b[w+])
         {
             w++; tmp++;
         }
     //    printf("%d %d\n",t,w);
         ans=max(ans,tmp);
     }
 //    for(int i=1;i<=m;i++) printf("%c",b[i]);
     printf("%d\n",ans);
     return ;
 }