CF526D Om Nom and Necklace

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我们可以把AB看成S，则要找的串可以写成SSSSA或者SSSSS。假设S出现了Q次，那么A出现了Q % k次，则B出现了 Q / k - Q % k次.

当ABABA是SSS的形式时，B可以为空字符，判断Q / k - Q % k>=0。

当ABABA是SSA的形式时，判断Q / k - Q % k > 0。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<stack>
 #include<queue>
 #include<vector>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e6 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), las = ' ';
   while(!isdigit(ch)) las = ch, ch = getchar();
   while(isdigit(ch)) ans = ans *  + ch - '', ch = getchar();
   if(las == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) putchar('-'), x = -x;
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, k;
 char s[maxn];
 int f[maxn];

 void init()
 {
   for(int i = , j = ; i <= n; ++i)
     {
       while(j && s[j + ] != s[i]) j = f[j];
       if(s[j + ] == s[i]) j++;
       f[i] = j;
     }
 }

 bool work(int i, int cir)
 {
   int x = i / cir;
   if(i % cir) return x / k > x % k;
   else return x / k >= x % k;
 }

 int main()
 {
   n = read(), k = read();
   scanf("%s", s + );
   init();
   for(int i = ; i <= n; ++i) write(work(i, i - f[i]));
   enter;
   return ;
 }