Good Triple

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    给出01串s，求数对[l,r]个数，使得能找到至少一对[x,k],使1<=x,k<=|s|且l<=x<x+2k<=r且s[x]=s[x+k]=s[x+2k]

Good Triple

Created by LXC on Tue Mar 5 13:19:54 2024

https://codeforces.com/problemset/problem/1168/B

ranting: 1900

tag: brute force, two pointers

problem

给出01串s，求数对[l,r]个数，使得能找到至少一对[x,k],使1<=x,k<=|s|且l<=x<x+2k<=r且s[x]=s[x+k]=s[x+2k]

solution

对于长度为9的串必有数对[l,r]。

我们考虑以i作为数对的第二个数共有多少数对，找到最大的k[i]，使得s[ k[i], i ]中能找到至少一对合法数对。

以i为右端点的贡献就是k[i]+1

code

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#include <bits/stdc++.h>
#define SINGLE_INPUT
#define ll long long
#define ull unsigned long long
#define N 500005
#define MOD 998244353
#define INF 0x3f3f3f3f
using namespace std;

template<class t,class u> ostream& operator<<(ostream& os,const pair<t,u>& p) {
    return os<<'['<<p.first<<", "<<p.second<<']';
}
template<class t> ostream& operator<<(ostream& os,const vector<t>& v) {
    os<<'['; int s = 1;
    for(auto e:v) { if (s) s = 0; else os << ", "; os << e; }
    return os<<']';
}
template<class t,class u> ostream& operator<<(ostream& os,const map<t,u>& mp){
    os<<'{'; int s = 1;
    for(auto [x,y]:mp) { if (s) s = 0; else os << ", "; os<<x<<": "<<y; }
    return os<<'}';
}

void sol() {
    string s;
    cin >> s;
    ll n = s.size();
    ll ans = 0;
    vector<ll> k(n, -1);
    for (ll i=1; i<n; i++) {
        ll j;
        k[i] = k[i-1];
        for (j=1; i-2*j>k[i]; j++) {
            if (s[i] == s[i-j] && s[i] == s[i-2*j]) {
                k[i] = i-2*j;
                break;
            }
        }
        ans += k[i]+1;
    }
    cout << ans << "\n";
}

int main() {
    cout << setprecision(15) << fixed;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifndef SINGLE_INPUT
    int t;
    cin >> t;
    while (t--) {
        sol();
    }
#else
    sol();
#endif
    return 0;
}