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  • Codeforces Round #565 (Div. 3) C. Lose it!

    链接:

    https://codeforces.com/contest/1176/problem/C

    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 <bits/stdc++.h>
    
    using namespace std;
    
    typedef long long LL;
    const int MAXN = 3e5 + 10;
    const int MOD = 1e9 + 7;
    int n, m, k, t;
    
    map<int, int> 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/YDDDD/p/10998734.html
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