标签：force   nim   possible   nsis   mod   amp   一个   called   def   

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;
}

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

标签：force   nim   possible   nsis   mod   amp   一个   called   def   

原文地址：https://www.cnblogs.com/YDDDD/p/10998734.html