BZOJ 3620 似乎在梦中见过的样子

KMP

显然给出的东西看起来就像border，因此有一个n^2的暴力KMP做法。我以为过不了，脑补了半天也没什么想法，然后发现题解就是n^2暴力？？？

#include<cstdio>
#include<cstring>
#define N 15555
using namespace std;
namespace runzhe2000
{
    char s[N];
    int n, k, next[N], ans;
    void main()
    {
        scanf("%s%d",s+1,&k); n = strlen(s+1);
        for(int i = 1; i <= n; i++)
        {
            int p = next[i] = i-1;
            for(int j = i+1; j <= n; j++)
            {
                for(; s[p+1] != s[j] && p >= i; p = next[p]);
                if(s[p+1] == s[j]) ++p;
                next[j] = p;
            }
            p = i-1;
            for(int j = i+1; j <= n; j++)
            {
                for(; s[p+1] != s[j] && p >= i; p = next[p]);
                if(s[p+1] == s[j]) ++p;
                for(; i+j-2*p-1 < 1 && p >= i; p = next[p]);
                if(p >= i+k-1) ans++;
            }
        }
        printf("%d\n",ans);
    }
}
int main()
{
    runzhe2000::main();
}