Niu Mei and 01 Chuan

// Problem:  Niu Meihe 01 strand 
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1043
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-03-01 19:24:46
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
int cnt[100005];
void solve() {
	string s;
    cin >> s;
    int n = s.size();
  
    for (int i = 0; i < n ;i ++)
    {
        cnt[i + 1] = s[i] == '0'?0 : 1;
        cnt[i + 1] += cnt[ i ];
    }
    int m = 0;
    for (int i = 1, l = 0; i <= n ; i++) {
        if (cnt[i] - cnt[l] && i - l > cnt[i] - cnt[l]) {
            l = i;
            m ++;
        }
        
    }
    cout << m << endl;
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

set up l Is the rightmost end of the previous segment . Then the properties that each segment needs to satisfy are 1 The number and 0 Are not equal and exist . Here, just use prefix and processing