传送门

题意：

思路：
一个同于最短路的板子题，初始的时候值为0，所以设$dis[0]=0$，让后选择一个最小的$a[i]$作为$base$，跑一遍同余最短路就好啦。跑完$dis[i]$表示在模$base$的意义下能到达$i$的最小数。让后问的是区间里的，转化为$[0,r]-[0,l-1]$就好啦。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n; LL l,r;
int a[N];
LL dis[N];
bool st[N];void dijkstra()
{memset(dis,INF,sizeof(dis));queue<int>q; q.push(0);dis[0]=0;while(q.size()){int u=q.front(); q.pop(); st[u]=false;for(int i=1;i<=n;i++){int now=(a[i]+u)%a[1];if(dis[now]>dis[u]+a[i]){dis[now]=dis[u]+a[i];if(!st[now]) q.push(now),st[now]=true;}}}
}LL get(LL x)
{LL sum=0;for(int i=0;i<a[1];i++) if(dis[i]<=x) sum+=(x-dis[i])/a[1]+1;return sum;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d%lld%lld",&n,&l,&r);for(int i=1;i<=n;i++) scanf("%d",&a[i]);dijkstra();printf("%lld\n",get(r)-get(l-1));return 0;
}
/**/