【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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<string>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<vector>
11 using namespace std;
12 typedef long long ll;
13 typedef unsigned int uint;
14 typedef unsigned long long ull;
15 typedef pair<int,int> PII;
16 typedef vector<int> VI;
17 #define fi first
18 #define se second 
19 #define MP make_pair
20 #define N   2100000
21 #define MOD 1000000007
22 #define eps 1e-8 
23 #define pi acos(-1)
24 #define oo 110000000000000
25 
26 
27 char a[N],b[N];
28 
29 
30 int read()
31 { 
32    int v=0,f=1;
33    char c=getchar();
34    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
35    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
36    return v*f;
37 }
38 
39 
40 void swap(int &x,int &y)
41 {
42     int t=x;x=y;y=t;
43 }
44 
45 
46 ull gcd(ull a, ull b)
47 {
48   return a == 0 ? b : gcd(b % a, a); 
49 } 
50 
51 
52 int main()
53 {
54 //    freopen("1.in","r",stdin);
55 //    freopen("1.out","w",stdout);
56     scanf("%s",a);
57     int n=strlen(a);
58     for(int i=1;i<=n;i++) b[i]=a[i-1];
59     for(int i=n+1;i<=n*2-1;i++) b[i]=b[i-n];
60     int m=n*2-1;
61     int t=0; 
62     int w=1;
63     int ans=1;
64     while(t<n)
65     {
66         t++; 
67         if(w<t) w=t;
68         int tmp=1;
69         while(w+1<=m&&w+1<=t+n-1&&b[w]!=b[w+1])
70         {
71             w++; tmp++;
72         }
73     //    printf("%d %d\n",t,w);
74         ans=max(ans,tmp);
75     }
76 //    for(int i=1;i<=m;i++) printf("%c",b[i]);
77     printf("%d\n",ans);
78     return 0;
79 }