Educational Codeforces Round 137 (Rated for Div. 2)

D. Problem with Random Tests

https://codeforces.com/contest/1743/problem/D

4 seconds / 512 megabytes

standard input / standard output

Problem

You are given a string $s$ consisting of $n$ characters. Each character of $s$ is either 0 or 1.

A substring of $s$ is a contiguous subsequence of its characters.

You have to choose two substrings of $s$ (possibly intersecting, possibly the same, possibly non-intersecting — just any two substrings). After choosing them, you calculate the value of the chosen pair of substrings as follows:

  • let $s_1$ be the first substring, $s_2$ be the second chosen substring, and $f(s_i)$ be the integer such that $s_i$ is its binary representation (for example, if $s_i$ is $\mathtt{11010}$, $f(s_i)$ = 26);
  • the value is the bitwise OR of $f(s_1)$ and $f(s_2)$.

Calculate the maximum possible value you can get, and print it in binary representation without leading zeroes.

Input

The first line contains one integer $n$ — the number of characters in $s$.

The second line contains $s$ itself, consisting of exactly $n$ characters $\mathtt{0}$ and/or $\mathtt{1}$.

All non-example tests in this problem are generated randomly: every character of $s$ is chosen independently of other characters; for each character, the probability of it being $\mathtt{1}$ is exactly $\frac{1}{2}$.

This problem has exactly $40$ tests. Tests from $1$ to $3$ are the examples; tests from $4$ to $40$ are generated randomly. In tests from $4$ to $10$, $n = 5$; in tests from $11$ to $20$, $n = 1000$; in tests from $21$ to $40$, $n = 10^6$.

Hacks are forbidden in this problem.

Output

Print the maximum possible value you can get in binary representation without leading zeroes.

Examples

input 1

5
11010

output 1

11111

input 2

7
1110010

output 2

1111110

input 3

4
0000

output 3

0

Note

In the first example, you can choose the substrings $\mathtt{11010}$ and $\mathtt{101}$. $f(s_1) = 26$, $f(s_2) = 5$, their bitwise OR is $31$, and the binary representation of $31$ is $\mathtt{11111}$.

In the second example, you can choose the substrings $\mathtt{1110010}$ and $\mathtt{11100}$.

题解

(下方字符串下标均从第 $1$ 位开始计,写程序时注意调整)

选择的子串之一一定是原串本身:

若从左到右第一个 $\mathtt{1}$ 为第 $x$ 位,答案最大位数为 $n-x+1$ 位。当且仅当选择原串时,能达到最大位数。

由此可知,其中一个子串选择原串时,为最好的情况。

暴力做法:

根据以上推导,我们已经可以得出一种做法,即其中一个子串选择原串,另一个子串选择原串第 $1$ 位到第 $i$ 位($i=1,2,\dots,n$),共 $n$ 种情况。另外,需要遍历整个字符串来得出两字符串做位运算的结果,每遍历一遍需要 $n$ 次循环。

子串都从第 $1$ 位开始取的原因:将选择原串的第 $x$ 位到第 $i$ 位,改为选择第 $1$ 位到第 $i$ 位,结果不会变差(只会不变或变好)。

综上,该暴力做法的时间复杂度为:$O(n^2)$,在 $10^6$ 的规模下一定会超时。

加速上述做法:

我们可以推断,为了最好的结果,我们选择的子串的起始点一定在第一个连续 $\mathtt{1}$ 区间内,并且这个子串的第一位一定填补了原串的第一个连续 $\mathtt{1}$ 区间后的第一个 $\mathtt{0}$ 位。

因为选择的子串只能填补子串起始点右侧的 $\mathtt{0}$,而越靠左的 $\mathtt{0}$ 位数越高,应当优先填充。如此,我们的子串起始点一定在第一个连续的 $\mathtt{1}$ 区间内。且起始点刚好能填补掉右侧的第一个 $\mathtt{0}$,设它为第 $x$ 位,为了满足这个条件,选择的子串长度为 $n-x+1$.

举一个例子演示上述思路:原串 $\mathtt{01101001}$

我们理应填充处于第 $1$ 位的 $\mathtt{0}$,但是由于它的左侧没有 $\mathtt{1}$,我们没办法填充。因此我们能填充的为第 $4$ 位的 $\mathtt{0}$,我们选择的子串起始点在第一个 $\mathtt{1}$ 区间内,即第 $2$ 位到第 $3$ 位。选择的子串长度为 $8-4+1=5$,这样刚好能覆盖掉第 $4$ 位的 $\mathtt{0}$.

即选择子串为 $\mathtt{11010}$ 或 $\mathtt{10100}$,至于具体选哪个,我们遍历第一个连续 $\mathtt{1}$ 区间的所有情况,找到最优的情况即可。

能直接遍历的原因:

上面的思路仍然需要遍历所有情况,但是这个情况数目就不是 $n$ 了,而是第一个连续 $\mathtt{1}$ 区间的长度。

由题意可知,数据为随机生成,每一位为 $0$ 和 $1$ 的概率均为 $1/2$ 。那么连续 $\mathtt{1}$ 区间的长度 $X$ 的期望为:

$$ E(X)=\sum_{i=1}^{n}{\frac{n}{2^n}}=2 $$

那么时间复杂度便为 $O(2n)$,非常合理。

代码

#include <bits/stdc++.h>

using namespace std;

string str_or(string &a, string &b)
{
    if (a.size() >= b.size())
    {
        string ans = a;
        for (int i = 0; i < b.size(); i++)
            if (b[i] == '1')
                ans[a.size() - b.size() + i] = '1';
        return ans;
    }
    return str_or(b, a);
}

string remove_zero(string &s)
{
    string ans = s;
    while (ans[0] == '0' && ans.size() - 1)
        ans.erase(ans.begin());
    return ans;
}

int main()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    int one = s.find('1'), zero = s.find('0', one);
    string ans = s;
    if (zero != s.npos)
    {
        for (int i = one; i < zero; i++)
        {
            string sub_s = s.substr(i, n - zero);
            ans = max(ans, str_or(s, sub_s));
        }
    }
    cout << remove_zero(ans) << endl;
    return 0;
}

标签: 数学, 字符串

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