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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#73547#4398. Luxury cruise shippoiAC ✓3ms5044kbC++1.3kb2023-01-25 18:21:202023-01-25 18:21:20

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.
  • [2023-01-25 18:21:20]
  • 评测
  • 测评结果:AC
  • 用时:3ms
  • 内存:5044kb
  • [2023-01-25 18:21:20]
  • 提交

answer

#include "iostream"
#include "cstring"
#include "cstdio"
#include "algorithm"
#include "queue"
#include "vector"
#include "queue"
#include "stack"
#include "ctime"
#include "set"
#include "map"
#include "cmath"
using namespace std;
#define fi first
#define se second
#define vi vector<int>
#define pb push_back
#define eb emplace_back
#define pii pair<int,int>
#define mp make_pair
#define rep( i , a , b ) for( int i = (a) , i##end = b ; i <= i##end ; ++ i )
#define per( i , a , b ) for( int i = (a) , i##end = b ; i >= i##end ; -- i )
#define mem( a ) memset( a , 0 , sizeof (a) )
#define all( x ) x.begin() , x.end()
//#define int long long
typedef long long ll;
const int MAXN = 2e5 + 10;
ll n , m;

ll f[MAXN];

void solve() {
	scanf("%lld",&n);
	ll res = n / 365 , re = 0x3f3f3f3f3f3f3f3f;
	n %= 365;
	rep( i , 0 , 30 ) {
		re = min( re , res + f[n] );
		if( res > 0 ) n += 365 , res --;
		else break;
	}
	printf("%lld\n",( re > 1e18 ? -1 : re ));
}

signed main() {
	memset( f , 0x3f , sizeof f );
	f[0] = 0;
	rep( i , 1 , 12000 ) {
		if( i >= 7 ) f[i] = min( f[i] , f[i - 7] + 1 );
		if( i >= 31 ) f[i] = min( f[i] , f[i - 31] + 1 );
	}
//	freopen("in","r",stdin);
//	freopen("out","w",stdout);
	int T;cin >> T;while( T-- ) solve();
//	solve();
}

Details

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Test #1:

score: 100
Accepted
time: 3ms
memory: 5044kb

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