E

E - Elevator
http://codeforces.com/gym/241680/problem/E
同余最短路，从0~a-1中每一个i向(i+b)%a连一条权值为b的边，向(i+c)%a连一条权值为c的边，然后跑spfa最短路，此时d[i]表示达到x%a花费的最小的距离，这里放的只有b,c，（解释一下，b,c组合出的实际大小为x,因为x过大，数组存不下，所以表示为x%a
最后统计答案的时候ans+=1+(h-d[i])/a;当前只有b,c组合出的d[i]算一个答案,然后剩下可以用a来填充

//用最小的来做同余系比较快

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define INF 9187201950435737471
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 100010
#define For(i,a,b) for(long long i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
long long h;
long long a,b,c,ans;
long long d[1000010];
queue<long long>q;
bool vis[1000010];

struct node{
    long long v;
    long long n;
    node *next;
}*e[10000010];

void in(long long &x){
    long long y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(long long x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}

void push(long long x,long long y,long long v){
    node *p;
    p=new node();
    p->n=y;
    p->v=v;
    if(e[x]==0)
        e[x]=p;
    else{
        p->next=e[x]->next;
        e[x]->next=p;
    }
}

void spfa(){
     For(i,0,a)
         d[i]=INF;
    q.push(1%a);
    d[1%a]=1;
    while(!q.empty()){
        long long t=q.front();
        q.pop();
        vis[t]=true;

        for(node *i=e[t];i;i=i->next){
            if(d[i->n]>d[t]+i->v){
                d[i->n]=d[t]+i->v;
                if(!vis[i->n]){
                    q.push(i->n);
                    vis[i->n]=true;
                }
            }
        }
        vis[t]=false;
    }
}

int main(){
     freopen("elevator.in","r",stdin);
     freopen("elevator.out","w",stdout);
    in(h);
    in(a);in(b);in(c);
    if(a>b) swap(a,b);
    if(a>c) swap(a,c);
    For(i,0,a-1){
        push(i,(i+b)%a,b);
        push(i,(i+c)%a,c);
    }
    spfa();
    For(i,0,a-1)
        if(h>=d[i])
            ans+=1+(h-d[i])/a;
    o(ans);
    return 0;
}