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题解

出题人的做法被hack了，不过幸亏他写了spj，数据可以保证是对的

这题的同余最短路做法是对的，而且还挺裸

关于同于最短路看这里

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
ll dis[maxn], p;
pll ans[maxn];
void dijkstra(vector<ll> v)
{
    priority_queue< pair<ll,ll>, vector< pair<ll,ll> >, greater< pair<ll,ll> > > heap;
    ll i;
    rep(i,1,p-1)dis[i]=linf;
    heap.em( make_pair( dis[0], 0 ) );
    ans[0] = {0,0};
    while(!heap.empty())
    {
        auto pr = heap.top(); heap.pop();
        if(pr.fi>dis[pr.se])continue;
        for(int i=0;i<v.size();i++)
        {
            auto to = (pr.se+v[i])%p;
            if(dis[to]>pr.fi+v[i])
            {
                dis[to] = pr.fi+v[i];
                ans[to] = {ans[pr.se].fi+!i,ans[pr.se].se+i};
                heap.em( make_pair(dis[to],to) );
            }
        }
    }
}
int main()
{
    ll a, b, c, k;
    cin >> a >> b >> c >> k;
    p = a;
    dijkstra({b,c});
    ll x, y=ans[k%a].fi, z=ans[k%a].se;
    cout << (k-b*y-c*z)/a << ' ' << y << ' ' << z;
    return 0;
}