51NOD 1554 欧姆诺姆和项链 巧妙利用KMP

请戳这里!

 #include<cstdio>
 #define N 1000100
 char s[N];
 int n,k,nxt[N],ans[N];
 int main()
 {
     scanf("%d%d%s",&n,&k,s+);
     for (int i=,j=;i<=n;i++)
     {
     while (j> && s[j+]!=s[i]) j=nxt[j];
     if (s[j+]==s[i]) j++;
     nxt[i]=j;
     }
     for (int i=;i<=n;i++)
     {
     int cnt=i/(i-nxt[i]);
     if (i%(i-nxt[i])==)
         if (cnt/k>=cnt%k)
         ans[i]=;
         else;
     else
         if (cnt/k>cnt%k)
         ans[i]=;
     }
     for (int i=;i<=n;i++)
     putchar(ans[i]+'');
     return ;
 }