题目链接

https://ac.nowcoder.com/acm/contest/8827/F

题意

给出一定数量的火柴棍，将这些火柴棍按照七段码形式摆放组成数字 A
问满足以上情况的最大数字 A 称为魔法数字

思路

直接暴力枚举所有满足情况的数字 A ，随着k的增大，魔法数字的总数是不断减小的。以下证明

代码

#include <bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", x)
#define mem(x, y) memset(x, y, sizeof(x))
#define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define pt putchar(10)
#define For(i, j, k) for (int i = (int)(j); i <= (int)(k); i++)
#define Rep(i, j, k) for (int i = (int)(j); i >= (int)(k); i--)
using namespace std;
typedef __int128 ll;
const int N = 1e5 + 7;
const ll mod = 1e9 + 7;
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
ll m, n, i, j;
int c[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
ll ans[10010];
inline void write(__int128 x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void dfs(int x, ll now, string s, int cost) {
    int i, j;
    if (x > 0)
        ans[cost] = max(ans[cost], now);
    int start = 0;
    if (x == 0)
        start = 1;
    for (i = start; i <= 9; i++) {
        if ((now * 10 + i) % (x + 1) == 0) {
            char ch = i + '0';
            dfs(x + 1, now * 10 + i, s + ch, cost + c[i]);
        }
    }
    return;
}
string t;
int main() {
    mem(ans, -1);
    dfs(0, 0, "", 0);
    cin >> t;
    if (t.size() > 7)
        cout << "-1" << endl;
    else {
        n = atoi(t.c_str());
        write(ans[n]);
    }
}