ZROI#960

先说$\Theta(n^2)$暴力叭.
显然的想法就是枚举答案中相邻两个$1$之间的$0$的个数.
然后贪心匹配,注意判断最后一段是否满足限制.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 1e5 + 100 ;

char s[N] ;
int n , ans , tmp[N] ;
bool v[N] ;

inline bool check () {

    return true ;
}

signed main (int argc , char * argv[]) {
    scanf ("%s" , s + 1 ) ; n = strlen ( s + 1 ) ;
    rep ( i , 1 , n ) v[i] = ( s[i] == '1' ) ;
    rep ( i , 0 , n ) {
        int time = 0 , res = 0 , cur = 0 ;
        bool f = false , d = 0 , able = false ;
        rep ( j , 1 , n ) {
            if ( ! d && ! v[j] ) {
                ++ cur ;
                if ( cur == i ) d = true , ++ time ;
            }
            if ( d && v[j] ) {
                cur = 0 ; d = false ;
                res += ( i + 1 ) ;
                able = true ;
            }
        }
        if ( able ) {
            if ( time < 2 ) continue ;
            if ( d ) { ans = max ( ans , res + i ) ; }
            else ans = max ( ans , res - 1 ) ;
        }
    }
    printf ("%lld\n" , ans ) ;
    return 0 ;
}

然后我们发现这个算法的瓶颈在于去匹配一串$0$,这里如果你对根号分治熟悉的话就可能会自然地想根号分治,这是完全可行的.
但我们要说的不是这种做法,我们考虑,虽然原序列中$0$的个数不存在单调性,但是前缀$0$的个数确实具有单调性的.
而且只需要关注以$1$为界的划分即可.于是我们就可以去二分/倍增地找一段$0$.
代码没写.