大意: 给定01字符串, 求有多少个区间$[l,r]$, 使得存在正整数$x,k$满足$1\le x,k\le n,l\le x<x+2k\le r,s_x=s_{x+k}=s_{x+2k}$.

 

0和1分开考虑, 那么问题就转化为给定排列求有多少个区间包含等差子序列, 可以用CF 452F Permutation的方法求出最小等差子序列位置, 然后就很容易能统计出答案.

复杂度是$O(n)$的

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+10;
int n, f[N];
char s[N];
vector<int> g[2];

int main() {
	scanf("%s", s+1);
	n = strlen(s+1);
	REP(i,1,n) g[s[i]-'0'].pb(i);
	REP(i,0,1) if (g[i].size()>=3) {
		int sz = g[i].size();
		REP(j,1,sz-2) {
			for (int k=j+1; k<min(sz,j+5); ++k) {
				int r = 2*g[i][k]-g[i][j], l = 2*g[i][j]-g[i][k];
				if (1<=r&&r<=n&&s[r]==s[g[i][j]]) { 
					f[r] = max(f[r], g[i][j]);
				}
				if (1<=l&&l<=n&&s[l]==s[g[i][j]]) {
					f[g[i][k]] = max(f[g[i][k]], l);
				}
			}
		}
	}
	REP(i,1,n) f[i] = max(f[i-1], f[i]);
	ll ans = 0;
	REP(i,1,n) ans += f[i];
	printf("%lld\n", ans);
}