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
#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (b); i >= (a); i--)
#define SZ(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define st first
#define nd second
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
auto &operator<<(auto &o, pair<auto, auto> p) {
    return o << "(" << p.st << ", " << p.nd << ")";
}
auto operator<<(auto &o, auto x)->decltype(end(x), o) {
    o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
    return o << "}";
}
#define deb(x...) cerr << "[" #x "]: ", [](auto...$) { ((cerr<<$<<"; "),...) << endl; }(x)

const int nax = 7777;
int n, k, m;
vector<pii> kolor[nax];
ll dp[nax][nax];
const ll inf = 1e18;

void upd(ll &x, ll y){
    x = min(x, y);
}

ll dist[nax];
bool vis[nax];

void solve(){
    cin >> n >> k >> m;
    rep(i, 1, n){
        int c; cin >> c;
        int wei, cost; cin >> wei >> cost;
        kolor[c].pb({wei, cost});
    }
    rep(i, 0, k){
        rep(j, 0, nax - 1) dp[i][j] = inf;
    }
    dp[0][0] = 0;
    rep(i, 0, k - 1){
        for(auto [wei, cost] : kolor[i + 1]){
            for(int rem=0;rem<m;rem++){
                upd(dp[i + 1][(rem + wei) % m], dp[i][rem] + cost);
            }
        }
    }

    rep(i, 0, m - 1) dist[i] = inf;
    dist[0] = 0;
    rep(runda, 1, m){
        ll mincost = inf + inf;
        int id = -1;
        rep(i, 0, m - 1){
            if(!vis[i] && dist[i] < mincost){
                mincost = dist[i];
                id = i;
            }
        }
        vis[id] = true;
        rep(i, 0, m - 1){
            upd(dist[(id + i) % m], dist[id] + dp[k][i]);
        }
    }

    rep(i, 0, m - 1){
        ll cur = dist[i];
        if(cur == inf) cur = -1;
        cout << cur << "\n";
    }

}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt = 1;
    // cin >> tt;
    rep(te, 1, tt) solve();
    return 0;
}