Problem J，Problem

Problem J，Problem

You’ll be given a string S​ and an integer K​. You have to find the number of non
super-boring substring inside S​. A substring is called super-boring if it contains any
palindromic substring of length ≥ K​ inside it.
For example, if S​ = “ababba” and K​ = 3​, then the substring “abab” is a super-boring
substring because it contains a palindrome “aba” whose length is ≥ 3. But the substring
“bba” is a valid non super-boring substring.
Find the number of substrings which are not super-boring. Two substrings are considered
different only if their starting or ending positions are different.
Input Specification
The first line of input contains an integer T​, denoting the number of test cases. The first
line of each test cases contains an integer K​ and the next line contains the string S
consisting of lower case English alphabets.
Constraints
1 ≤ T ≤ 100
K ≤ |S| ≤ 10
5
Sum of |S| over all test cases <= 10
6
Note: Here |S| denotes the length of the string S.
Output Specification
For each test case you have to print the number of non super-boring substring of the
string S​.
Statements
Input Output
3
3
bababe
4
abcbbcbc
3
aabacecc
12
27
20

题意：

给你一个字符串，求区间内不包含长度k以上的回文串的子串个数

题解：

马拉车先求出回文长度，之后每个字符找过去，易证马拉车的回文半径-1就是字符串的回文长度，然后只需要找符合k的最短长度，如果是不同奇偶的话就是+1,那么在这个字符串前面有(i+1)/2-len/2个字符，这些字符就是包括长度至少为k的子串，那么我们就把他加到树状数组中，然后树状数组记录的是最大值，就代表以pos为结尾的子串，含有长度为k的回文串的子串有多少个
举个例子：
假设有asdfgfzxc这么一个字符串，k=3，很明显fgf是满足条件的，那么在后一个f之后便放到树状数组中4代表从6这个位置开始有4个满足要求的回文串,z,x,c这些位置都要减4.

#include<bits/stdc++.h>
using namespace std;
#define maxn 1000050
#define ll long long
char s[maxn];
char ss[2*maxn];
int p[2*maxn];
void manacher(char s[],int len)
{
    int i,j,len1;
    for(i=0;i<2*len+2;i++)ss[i]='#';
    for(i=0;i<len;i++)ss[i*2+2]=s[i];
    len1=len*2+1;ss[0]='$';
    int mx=0,id=0;
    int ans=0;
    for(int i=0;i<len1;i++){
        p[i]=mx>i?min(p[2*id-i],mx-i ):1;
        while(ss[i+p[i] ]==ss[i-p[i] ] )p[i]++;
        if(ans<p[i])ans=p[i];
        if(i+p[i]>mx){
            mx=i+p[i];
            id=i;
        }
    }
}
int sum[maxn];
int mlen;
int lowbit(int x)
{
    return x&(-x);
}
void add(int pos,int val)
{
    for(int i=pos;i<=mlen;i+=lowbit(i))
        sum[i]=max(sum[i],val);
}
ll query(int pos)
{
    ll ans=0;
    for(int i=pos;i>=1;i-=lowbit(i))
        ans=max(ans,(ll)sum[i]);
    return ans;
}
vector<int>vec[500005];
int len[500005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        int k;
        scanf("%d",&k);
        scanf("%s",s);
        manacher(s,strlen(s));
        mlen=strlen(ss);
        ll ans=0;
        for(int i=2;i<=mlen;i++)
        {
            int len=p[i]-1;
            if(len<k)
                continue;
            if((len+k)%2==0)
                len=k;
            else
                len=k+1;
            int val=(i+1)/2-len/2;
            int pos=val+len-1;
            add(pos,val);
        }
        int n=strlen(s);
        for(int i=1;i<=n;i++)
        {
            int num=query(i);
            ans+=i-num;
        }
        printf("%lld\n",ans);
    }
    return 0;
}