求改USACO铜组T2

//
// Created by 55062 on 2023/12/17.
//
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node{
    ll l,r,zx,shu,zt;
}ans[300010];
ll n,cnt,x,sum,l=1,abc=0x3f3f3f3f,abb;
char a[300010];
int main(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    a[n+1]='0';
    for(int i=1;i<=n;i++)
        if(a[i]=='1') sum++;
    for(int i=1;i<=n+1;i++) {
        if (a[i] == '1') x++;
        if (a[i] == '1' && a[i + 1] == '0') {
            ans[++cnt].shu = x;
            ans[cnt].l=l;
            ans[cnt].r=i;
            l=i+1;
            x=0;
        }if(a[i]=='0') l++;
    }/*
    for(int i=1;i<=cnt;i++)
        if(ans[i].shu==1||(ans[i].shu%2==0&&ans[i].l!=1&&ans[i].r!=n)){
            cout<<sum;
            exit(0);
        }*/
    for(int i=1;i<=cnt;i++){
        if(ans[i].shu%2==0) ans[i].zx=ans[i].shu/2;
        else ans[i].zx=(ans[i].shu-1)/2;
    }for(int i=1;i<=cnt;i++)
        if(ans[i].zx!=0) abc=min(abc,ans[i].zx);
    for(int i=1;i<=cnt;i++){
        if(ans[i].zx==abc) ans[i].zt=1;
        else if(ans[i].l!=1&&ans[i].r!=cnt){
            ll l=ans[i].l+abc*2,r=ans[i].r-abc*2;
            ans[i].zt=2;
            ll x=r-l-1;
            if(x>0){
                ll h=abc*2+1;
                ans[i].zt+=x/h;
                if(x%h!=0) ans[i].zt++;
            }
        }else{
            ll h=abc*2+1;
            ans[i].zt=ans[i].shu/h;
            if(ans[i].zt%h!=0) ans[i].zt++;
        }
    }for(int i=1;i<=cnt;i++) abb+=ans[i].zt;
    cout<<abb;
    return 0;
}