目录

CF1463F-Max Correct Set

CF1463F-Max Correct Set

题目:

题目描述:

Let’s call the set of positive integers $ S $ correct if the following two conditions are met:

  • $ S \subseteq {1, 2, \dots, n} $ ;
  • if $ a \in S $ and $ b \in S $ , then $ |a-b| \neq x $ and $ |a-b| \neq y $ .

For the given values $ n $ , $ x $ , and $ y $ , you have to find the maximum size of the correct set.

输入格式:

A single line contains three integers $ n $ , $ x $ and $ y $ ( $ 1 \le n \le 10^9 $ ; $ 1 \le x, y \le 22 $ ).

输出格式:

Print one integer — the maximum size of the correct set.

样例:

样例输入1:

1
10 2 5

样例输出1:

1
5

样例输入2:

1
21 4 6

样例输出2:

1
9

样例输入3:

1
1337 7 7

样例输出3:

1
672

样例输入4:

1
455678451 22 17

样例输出4:

1
221997195

思路:

实现:

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// Problem: CF1463F Max Correct Set
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF1463F
// Memory Limit: 250 MB
// Time Limit: 4000 ms
// Author: Ybw051114
//
// Powered by CP Editor (https://cpeditor.org)

#include "ybwhead/ios.h"

const int ImaxnF = 1e9;
const int maxn = 22;

int dp[2][1 << maxn];
int val[2 * maxn];

int main()
{
    int n, x, y;
    yin >> n >> x >> y;

    int k = x + y;
    int m = max(x, y);
    int FULL = (1 << m) - 1;

    for (int i = 0; i < k; ++i)
        val[i] = n / k + (i < n % k);
    for (int mask = 0; mask < (1 << m); ++mask)
        dp[0][mask] = -ImaxnF;
    dp[0][0] = 0;

    for (int i = 0; i < k; ++i)
    {
        for (int mask = 0; mask < (1 << m); ++mask)
            dp[1][mask] = -ImaxnF;
        for (int mask = 0; mask < (1 << m); ++mask)
        {
            if (dp[0][mask] == -ImaxnF)
                continue;
            int nmask = (mask << 1) & FULL;
            dp[1][nmask] = max(dp[1][nmask], dp[0][mask]);
            if (((mask >> (x - 1)) & 1) | ((mask >> (y - 1)) & 1))
                continue;
            nmask |= 1;
            dp[1][nmask] = max(dp[1][nmask], dp[0][mask] + val[i]);
        }
        swap(dp[0], dp[1]);
    }

    int ans = 0;
    for (int mask = 0; mask < (1 << m); ++mask)
        ans = max(ans, dp[0][mask]);
    yout << ans << endl;
}