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Codeforces Round #706 (Div. 2)-B. Max and Mex

传送门 Time Limit: 1 second Memory Limit: 256 megabytes

Problem Description

You are given a multiset $S$ initially consisting of $n$ distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.

You will perform the following operation $k$ times:

Here $\operatorname{max}$ of a multiset denotes the maximum integer in the multiset, and $\operatorname{mex}$ of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:

Your task is to calculate the number of distinct elements in $S$ after $k$ operations will be done.

Input

The input consists of multiple test cases. The first line contains a single integer $t$ ( $1\le t\le 100$ ) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ , $k$ ( $1\le n\le 10^5$ , $0\le k\le 10^9$ ) — the initial size of the multiset $S$ and how many operations you need to perform.

The second line of each test case contains $n$ distinct integers $a_1,a_2,\dots,a_n$ ( $0\le a_i\le 10^9$ ) — the numbers in the initial multiset.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$ .

Output

For each test case, print the number of distinct elements in $S$ after $k$ operations will be done.

Sample Input

Sample Onput

Note

In the first test case, $S=\{0,1,3,4\}$ , $a=\operatorname{mex}(S)=2$ , $b=\max(S)=4$ , $\lceil\frac{a+b}{2}\rceil=3$ . So $3$ is added into $S$ , and $S$ becomes $\{0,1,3,3,4\}$ . The answer is $4$ .

In the second test case, $S=\{0,1,4\}$ , $a=\operatorname{mex}(S)=2$ , $b=\max(S)=4$ , $\lceil\frac{a+b}{2}\rceil=3$ . So $3$ is added into $S$ , and $S$ becomes $\{0,1,3,4\}$ . The answer is $4$ .

题目大意

MAX是一个集合中最大的数 MEX是一个集合中最小的非负整数每次操作把 $\lceil\frac{MAX+MEX}{2}\rceil$ 向上取整，得到的新数放入集合中。问进行k次操作后，集合中有几种数。

题目分析

考虑两种情况，初始时 MEX $\ge$ n 和 MEX $\le$ n 。

MEX $\ge$ n_{其实就是MEX=n}

读者不妨计算几个试试，将会发现每进行一次操作，MEX就会+1，集合中不同元素的个数也就随之+1。k次操作后，集合中不同元素的个数就变成了n + k

MEX $\le$ n

如果k=0，不需要进行操作，答案就是最初的集合中不同元素的个数

否则新生成的数Mnew一定是 $\ge$ MEX并且 $\le$ MAX的，每次操作都是这个数(因为新生成的数不会改变MEX和MAX)。

解题思路

读入n个a后，排序，遍历一遍得到MEX，并且用map去重。

若是MEX $\ge$ n，就输出n+k

否则

如果k=0，就直接输出map.size

否则，计算一次新生成的数Mnew，并将Mnew插入到map中，输出map.size

_{map可以设置为map<int,bool>，map.size就是集合中不同元素的个数}

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int a[100010];

int main()
{
    int N;
    cin >> N;
    while (N--) //N个测试样例
    {
        int n, k;
        scanf("%d%d", &n, &k);
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]); //读入
        }
        sort(a, a + n); //排序
        int Me = 0;
        map<int, bool> mp;
        for (int i = 0; i < n; i++)
        {
            mp[a[i]] = 1; //a[i]出现过，插入map
            if (a[i] == Me) //如果目前的Me出现过，就Me++
                Me++;
        }
        if (Me >= n) //第一种情况，直接输出
        {
            printf("%d\n", n + k);
            continue;
        }
        else //否则
        {
            map<int, bool>::iterator it = mp.end();
            it--;
            int Ma = it->first; //Ma就是map中的最后一个的first
            int Mnew = (Ma + Me) / 2 + ((Ma + Me) % 2 != 0); //操作后新生成的数
            if (k != 0) //k不等于0
                mp[Mnew] = 1;
            printf("%d\n", mp.size());
        }
    }
    return 0;
}

同步发文于我的CSDN,原创不易，转载请附上原文链接哦~
Tisfy：blog.csdn.net/Tisfy/artic…

Codeforces Round #706 (Div. 2)-B. Max and Mex-题解