【每日一题】Problem 538B. Quasi Binary

合集 - 每日一题(23)

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23.【每日一题】Problem 538B. Quasi Binary2023-07-19

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原题

解决思路

最简单的思路就是贪心了，每次生成不超过目标值的 $quasibinary$，即可使最终数量最少

#include <bits/stdc++.h>

int quasibinary(int max)
{
    int res = 0;
    int p = 0;
    while (max > 0) {
        if (max % 10 > 0) {
            res += int(pow(10, p));
        }
        max /= 10;
        ++p;
    }
    return res;
}

int main() 
{
    int n; std::cin >> n;
    std::vector<int> res;
    while (n > 0) {
        int target = quasibinary(n);
        n -= target;
        res.push_back(target);
    }

    std::cout << res.size() << std::endl;
    for (auto &v : res) {
        std::cout << v << " ";
    }
    std::cout << "\n";
    return 0;
}

其他解

可以考虑使用 dp，待补充
完全背包
$dp[i][j]$ 指代当目标值为 $j$ 时，当前值取 $i$ 个可以达到目的的最少数量