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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#73547 | #4398. Luxury cruise ship | poi | AC ✓ | 3ms | 5044kb | C++ | 1.3kb | 2023-01-25 18:21:20 | 2023-01-25 18:21:20 |
Judging History
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