Codeforces Round #565 (Div. 3) C. Lose it!
Codeforces Round #565 (Div. 3)
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 66 and it is possible to split it into k6k6 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 4,8,15,16,23,424,8,15,16,23,42);
- [4,8,15,16,23,42,4] (the length of the array is not divisible by 66);
- [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).
Input
The first line of the input contains one integer n (1≤n≤5⋅105) — the number of elements in a.
The second line of the input contains n integers a1,a2,…,ana1,a2,…,an (each ai is one of the following numbers: 4,8,15,16,23,424,8,15,16,23,42), where ai is the i-th element of aa.
Output
Print one integer — the minimum number of elements you have to remove to obtain a good array.
Examples
input
5
4 8 15 16 23
output
5
input
12
4 8 4 15 16 8 23 15 16 42 23 42
output
0
input
15
4 8 4 8 15 16 8 16 23 15 16 4 42 23 42
output
3
题意：题目把 [4,8,15,16,23,42] 定义为一个好的子序列，然后给你一个数组，你可以对数组进行无数次操作（删除一个元素），
问你最少操作多少次，可以将该数组变成好的序列（只要保证序列里的每个 [4,8,15,16,23,42] 序列的顺序就行了，相互之间可以交叉）
思路：我这里写法是用6个队列，分别存下数组中出现的每种数字[4,8,15,16,23,42]的下标，存完之后从第一个标记 4 数字 下标的队列开始遍历6个队列，每找出一个符合 好的子序列 顺序的六个下标，计数变量加一，关键来了，如果找的时候顺序不对，
例如 8 4 8 15 16 23 ，先找到 4 下标 1，找 8 时 出现 0 比 1 还小，说明这个 8 在序列中无效！！所以把8所在队列pop掉队首，
继续在8所在队列往后找，直到找到完整6个数，则把计数变量加一，6个队列都pop掉队首，
另外，循环结束条件显然是6个队列的某个队列为空，如果用个solve()函数写主体程序，可以直接return，
比较方便，优势也在这里，在main函数还得用标记退出几层循环，比较烦。
最后答案也就是输出总个数减去符合好的序列的数量乘以6......
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<cmath>
5 #include<algorithm>
6 #include<map>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 using namespace std;
11 #define ll long long
12 const int mod=1e9+7;
13 const int inf=1e9+7;
14
15 //const int maxn=;
16
17 queue<int>q[6];
18
19 ll int cnt=0;
20
21 void solve()
22 {
23 while(!q[0].empty())
24 {
25 int now=q[0].front();
26 for(int i=1;i<6;i++)
27 {
28 if(q[i].empty())
29 {
30 return ;
31 }
32 if(q[i].front()>now)
33 {
34 now=q[i].front();
35 continue;
36 }
37 else if(q[i].front()<now)
38 {
39 q[i].pop();
40 while(q[i].front()<now)
41 {
42 if(q[i].empty())
43 {
44 return ;
45 }
46 q[i].pop();
47 }
48 now=q[i].front();
49 }
50 }
51 cnt++;
52 for(int i=0;i<6;i++)
53 q[i].pop();
54 }
55 }
56 int main()
57 {
58 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
59 int n;
60 cin>>n;
61 int num;
62 for(int i=0;i<n;i++)
63 {
64 cin>>num;
65 if(num==4)
66 q[0].push(i);
67 else if(num==8)
68 q[1].push(i);
69 else if(num==15)
70 q[2].push(i);
71 else if(num==16)
72 q[3].push(i);
73 else if(num==23)
74 q[4].push(i);
75 else if(num==42)
76 q[5].push(i);
77 }
78
79 cnt=0;
80
81 solve();
82
83 cout<<n-6*cnt<<endl;
84
85 return 0;
86 }
