//#pragma GCC optimize("Ofast") //#pragma GCC optimize("unroll-loops") #include using namespace std; using ll = long long; using ull = unsigned long long; using lll = __int128_t; using ld = long double; using P = pair; using vi = vector; using vd = vector; using vP = vector

; using vS = vector; using vvi = vector; using vvd = vector; using vvP = vector; using v3i = vector; using v3d = vector; using v3P = vector; using v4i = vector; using v5i = vector; using v6i = vector; #if __has_include() #include using namespace atcoder; using mint = modint998244353; using Mint = modint1000000007; using vm = vector; using vM = vector; using vvm = vector; using vvM = vector; using v3m = vector; using v3M = vector; using v4m = vector; using v5m = vector; #endif template using rp_Q = priority_queue, greater>; #define rrep(i, n) for (ll i = (n)-1; (i) >= 0; --(i)) #define rep(i, n) for (ll i = 0; (i) < (n); ++(i)) #define loop(i, a) for(ll i = a; true; ++(i)) #define reps(i, n) for (ll i = 1; (i) <= (n); ++(i)) #define Rep(i,a,b) for(ll i = (a); i <= (b); i++) #define all(a) (a).begin(),(a).end() const ll MOD = 998244353; const ll Hash = 10000000000000061; #define Yes(b) ((b)?cout << "Yes" << endl:cout << "No" << endl) #define YES(b) ((b)?cout << "YES" << endl:cout << "NO" << endl) #define Aoki(b) ((b)?cout << "Aoki" << endl:cout << "Takahashi" << endl) #define out(x) cout << (x) << endl #define vout(v) for(ll x : (v)) cout << x << endl #define vin(v) rep(i, (v).size()) cin >> (v)[i] #define ft first #define sd second #define vInP(v) rep(i, (v).size()) cin >> (v)[i].ft >> (v)[i].sd bool chmin(ll& a, ll b) { return a > b ? a = b, 1 : 0; } bool chmax(ll& a, ll b) { return a < b ? a = b, 1 : 0; } vi s90 = { 0, 1, 0, -1 }, c90 = { 1, 0, -1, 0 }; vi s45 = { 0, 1, 1, 1, 0, -1, -1, -1 }, c45 = { 1, 1, 0, -1, -1, -1, 0, 1 }; void solve(); signed main(){ cin.tie(nullptr); ios_base::sync_with_stdio(false); cout << fixed << setprecision(20); int t = 1; // cin >> t; rep(i, t) solve(); } void solve(){ string S; cin >> S; reverse(all(S)); ll N = S.size(); vi a(N); rep(i, N) a[i] = S[i] - '0'; ll ans = 0, p = 0, P = 0; while(true){ if(p > N - 3) break; if((a[p] == 0) && (a[p + 1] == 1) && a[p + 2] == 1){ ans++; a[p] = 1; a[p+2] = 0; ll q = p - 1; while(q > p - 3){ if(q < 0) break; if((a[q] == 0) && (a[q + 1] == 1) && a[q + 2] == 1){ ans++; a[q] = 1, a[q + 2] = 0; p = q; } q--; } } p = P+1; P++; } out(ans); }