链接：

题意：

You are given an array a consisting of n integers. Each ai is one of the six following numbers: 4,8,15,16,23,42.

Your task is to remove the minimum number of elements to make this array good.

An array of length k is called good if k is divisible by 6 and it is possible to split it into k6 subsequences 4,8,15,16,23,42.

Examples of good arrays:

[4,8,15,16,23,42] (the whole array is a required sequence);

[4,8,4,15,16,8,23,15,16,42,23,42] (the first sequence is formed from first, second, fourth, fifth, seventh and tenth elements and the second one is formed from remaining elements);

[] (the empty array is good).

Examples of bad arrays:

[4,8,15,16,42,23] (the order of elements should be exactly 4,8,15,16,23,42);

[4,8,15,16,23,42,4] (the length of the array is not divisible by 6);

[4,8,15,16,23,42,4,8,15,16,23,23] (the first sequence can be formed from first six elements but the remaining array cannot form the required sequence).

思路：

刚开始题目看错。。以为任意顺序的子序列。

map将值映射到1-6，然后每次遇到2-5中的x，从x-1的个数中移一个到x，最后看能移几个到6.

代码：

#include 
    
     using namespace std;typedef long long LL;const int MAXN = 3e5 + 10;const int MOD = 1e9 + 7;int n, m, k, t;map
     
       mp;int main(){    cin >> n;    mp[4] = 1;    mp[8] = 2;    mp[15] = 3;    mp[16] = 4;    mp[23] = 5;    mp[42] = 6;    int vis[10] = {0};    int res = 0, temp = 0;    int last = 1;    int flag = 0;    for (int i = 1;i <= n;i++)    {        int v;        cin >> v;        if (mp[v] == 1)            vis[mp[v]]++;        else if (vis[mp[v]-1]>0)        {            vis[mp[v]-1]--;            vis[mp[v]]++;        }    }    cout << n-6*vis[6] << endl;    return 0;}