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赛前训练4 字符串哈希

news 2025/10/2 19:20:21

A

显然长度具有单调性，于是二分长度+map存储哈希值判断即可。

实现

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define ull unsigned long long
#define int long long
using namespace std;const int N=2e4+5;
const ull BASE=13331;
int n,k;
int a[N];
ull P[N],H[N];
string s;
map<ull,int> mp;int get(int l,int r){return H[r]-P[r-l+1]*H[l-1];
}
bool check(int x){for(int i=1;i+x-1<=n;i++){int h=get(i,i+x-1);mp[h]++;if(mp[h]==k)return 1;}return 0;
}signed main(){ios::sync_with_stdio(0);//cin.tie(0);cin>>n>>k;P[0]=1;for(int i=1;i<=n;i++)cin>>a[i],H[i]=H[i-1]*BASE+a[i],P[i]=P[i-1]*BASE;int l=0,r=n+1;while(l+1<r){int mid=(l+r)>>1;//cout<<mid<<'\n';if(check(mid))l=mid;elser=mid;}cout<<l;return 0;
}

B

把所有回文子串预处理出来然后树状数组统计即可。

实现

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define ull unsigned long long
#define int long long
#define pii pair<int,int>
using namespace std;const int N=2e3+5;
const ull BASE=13331;
int n,tot;
ull P[N],H1[N],H2[N];
pii p[N*N];
int tree[N];
string s;int geth1(int l,int r){return H1[r]-P[r-l+1]*H1[l-1];
}
int geth2(int l,int r){return H2[l]-P[r-l+1]*H2[r+1];
}
bool cmp(pii &x,pii &y){if(x.first==y.first)return x.second<y.second;return x.first<y.first;
}
int lowbit(int x){return x&-x;
}
void upd(int x,int y){for(;x<=n;x+=lowbit(x))tree[x]+=y;
}
int qry(int x){int res=0;for(;x;x-=lowbit(x))res+=tree[x];return res;
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cin>>s;n=s.size(),s='#'+s;P[0]=1;for(int i=1;i<=n;i++)P[i]=P[i-1]*BASE,H1[i]=H1[i-1]*BASE+s[i];for(int i=n;i>=1;i--)H2[i]=H2[i+1]*BASE+s[i];for(int i=1;i<=n;i++)for(int j=i;j<=n;j++)if(geth1(i,j)==geth2(i,j))p[++tot]={i,j};sort(p+1,p+tot+1,cmp);int ans=0;for(int i=tot;i>=1;i--){ans+=qry(n)-qry(p[i].second);upd(p[i].first,1);}cout<<ans;return 0;
}

C

删掉本来就回文的前后缀，剩下的变成删除前缀/后缀的问题，可以枚举删掉的长度然后二分最长回文前后缀的长度用哈希判断即可。

实现

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define int long long
#define ull unsigned long long
using namespace std;const int N=5e5+5;
const ull BASE=13331;
int n;
ull P[N],H1[N],H2[N];
string s;int geth1(int l,int r){return H1[r]-P[r-l+1]*H1[l-1];
}
int geth2(int l,int r){return H2[l]-P[r-l+1]*H2[r+1];
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cin>>s,n=s.size(),s='#'+s;P[0]=1;for(int i=1;i<=n;i++)P[i]=P[i-1]*BASE,H1[i]=H1[i-1]*BASE+s[i];for(int i=n;i>=1;i--)H2[i]=H2[i+1]*BASE+s[i];int l=1,r=n;while(l<r&&s[l]==s[r])l++,r--;int ans=l-1;if(l>=r){cout<<ans;return 0;}int extra=-1e18;for(int i=l;i<=r;i++){int lt=-1,rt=(r-i+1)/2+1;while(lt+1<rt){int mid=(lt+rt)>>1;if(geth1(i,i+mid-1)==geth2(r-mid+1,r))lt=mid;elsert=mid;}extra=max(extra,lt);}for(int i=r;i>=l;i--){int lt=-1,rt=(i-l+1)/2+1;while(lt+1<rt){int mid=(lt+rt)>>1;if(geth1(l,l+mid-1)==geth2(i-mid+1,i))lt=mid;elsert=mid;}extra=max(extra,lt);}cout<<ans+extra;return 0;
}

D

取反之后正反各做一遍哈希，然后注意到这个题的回文子串必定是偶数长度的，于是统计回文子串的数量即可。这是经典套路。（即枚举对称中心二分半径）

实现

//
//  C.cpp
//
//
//  Created by _XOFqwq on 2024/10/23.
//#include <bits/stdc++.h>
#define int long long
#define ull unsigned long long
using namespace std;const int N=5e5+5;
const int BASE=13331;
int n,ans;
ull hs1[N],hs2[N],p[N];
string s;void init_hs(){p[0]=1;for (int i=1; i<=n; i++) {hs1[i]=hs1[i-1]*BASE+s[i];p[i]=p[i-1]*BASE;}for (int i=n; i>=1; i--) {hs2[i]=hs2[i+1]*BASE+(s[i]=='0'?'1':'0');}
}
ull get_hs1(int l,int r){return hs1[r]-hs1[l-1]*p[r-l+1];
}
ull get_hs2(int l,int r){return hs2[l]-hs2[r+1]*p[r-l+1];
}
bool check(int pos,int x){int l=pos-x+1,r=pos+x;return get_hs1(l,r)==get_hs2(l,r);
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cin>>n>>s,s='#'+s;init_hs();for (int i=1; i<n; i++) {int l=0,r=min(i,n-i)+1;while (l+1<r) {int mid=(l+r)>>1;if (check(i,mid)) {l=mid;} else {r=mid;}}ans+=l;}cout<<ans;return 0;
}

总结：