Problem B. 小朋友的 mex 题目英文名：mex 时空限制：1s/256MB 题目描述： 函数 mex(a,b,c......)表示在 a,b,c......中最小的没有出现的自然数 。 有 n 个小朋友，每个小朋友有一个数，分别为a[1] 到 a[n] 。 小朋友们想求出： 对于 a 数组的所有连续子区间，求出它们的 mex的和。 由于答案的极限大小大约是 10e15 ，小朋友们就不要求你取模了。 （居然提醒大家开 long long ，多么良心的出题人！） C++代码实现

Sure! Here's a C++ implementation for the problem:

#include <iostream>
#include <vector>

using namespace std;

long long mex(vector<int>& a) {
    int n = a.size();
    vector<bool> visited(n+1, false);
    
    for (int num : a) {
        if (num >= 0 && num <= n) {
            visited[num] = true;
        }
    }
    
    for (int i = 0; i <= n; i++) {
        if (!visited[i]) {
            return i;
        }
    }
    
    return n + 1;
}

long long sumMex(vector<int>& a) {
    int n = a.size();
    long long sum = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            vector<int> subArray(a.begin() + i, a.begin() + j + 1);
            sum += mex(subArray);
        }
    }
    
    return sum;
}

int main() {
    int n;
    cin >> n;
    
    vector<int> a(n);
    
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    
    long long result = sumMex(a);
    
    cout << result << endl;
    
    return 0;
}

This code defines two functions: `mex` and `sumMex`. The `mex` function finds the minimum non-negative integer that is not present in the given array. The `sumMex` function calculates the sum of mex values for all the contiguous subarrays of the input array. The main function reads the input, calls `sumMex`, and prints the result.

Please note that this implementation assumes that the input array contains non-negative integers. If the array can contain negative integers as well, you may need to modify the code accordingly.

Let me know if you have any further questions!