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QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#45426#4398. Luxury cruise shipmiaomiaoziAC ✓2ms4396kbC++171.3kb2022-08-23 18:37:352022-08-23 18:37:37

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-08-23 18:37:37]
  • 评测
  • 测评结果:AC
  • 用时:2ms
  • 内存:4396kb
  • [2022-08-23 18:37:35]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
// https://space.bilibili.com/672346917

#ifndef LOCAL
#define LOG(...) 42
#endif

#define fi first
#define se second
#define pb push_back
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

typedef long long LL;
typedef pair <int, int> PII;

constexpr double EPS = 1e-8;
const double PI = acos(-1);

int multi_cases = 1;

constexpr int N = 7 * 31 * 365 * 2;
constexpr LL inf = 1e18;
int do_init = 0;
vector <LL> f(N + 5, inf);

void A_SOUL_AvA () {
    if (!do_init) {
        do_init = 1;
        f[7] = f[31] = f[365] = 1;
        vector <int> a = {7, 31, 365};
        for (int v : a) {
            for (int j = v; j <= N; j++) {
                f[j] = min(f[j], f[j - v] + 1);
            }
        }
    }

    LL n;
    cin >> n;

    LL k = N / 361;
    LL ans = 0;
    if (n <= N) {
        cout << (f[n] == inf ? -1 : f[n]) << endl;
    } else {
        LL x = n / N;
        n %= N;
        ans += x * f[N];

        cout << (f[n] == inf ? -1 : f[n] + ans) << endl;
    }
}

int main () {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(12);

    int T = 1;
    for (multi_cases && cin >> T; T; T--) {
        A_SOUL_AvA ();
    }

    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 4396kb

input:

1000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101...

output:

-1
-1
-1
-1
-1
-1
1
-1
-1
-1
-1
-1
-1
2
-1
-1
-1
-1
-1
-1
3
-1
-1
-1
-1
-1
-1
4
-1
-1
1
-1
-1
-1
5
-1
-1
2
-1
-1
-1
6
-1
-1
3
-1
-1
-1
7
-1
-1
4
-1
-1
-1
8
-1
-1
5
-1
-1
2
9
-1
-1
6
-1
-1
3
10
-1
-1
7
-1
-1
4
11
-1
-1
8
-1
-1
5
12
-1
-1
9
-1
-1
6
13
-1
3
10
-1
-1
7
14
-1
4
11
-1
-1
8
15
-1
5
12
-1
-...

result:

ok 1000 lines