思路：
经典题 不解释
找到最小的数mn
所有都是在mod mn的意义下 搞得
i->(i+a[i])%mn 边权为a[i]
//By SiriusRen #include <queue> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int N=500000*13,inf=0x3f3f3f3f; int n,first[N],next[N],v[N],tot,minn=inf,a[15],vis[N]; ll BMin,BMax,w[N],dis[N],ans; void add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;} struct Node{int now,dis;}st,jy; bool operator<(Node a,Node b){return a.dis>b.dis;} priority_queue<Node>pq; int main(){ memset(first,-1,sizeof(first)); memset(dis,0x3f,sizeof(dis)); scanf("%d%lld%lld",&n,&BMin,&BMax); for(int i=1;i<=n;i++)scanf("%d",&a[i]),minn=min(minn,a[i]); dis[0]=0; for(int i=1;i<=n;i++) for(int j=0;j<minn;j++) add(j,(j+a[i])%minn,a[i]); pq.push(st); while(!pq.empty()){ Node t=pq.top();pq.pop(); if(vis[t.now])continue; vis[t.now]=1; for(int i=first[t.now];~i;i=next[i]) if(dis[v[i]]>dis[t.now]+w[i]) dis[v[i]]=dis[t.now]+w[i], jy.now=v[i],jy.dis=dis[v[i]],pq.push(jy); }BMin--; for(int i=0;i<minn;i++){ if(dis[i]<=BMin)ans-=(BMin-dis[i])/minn+1; if(dis[i]<=BMax)ans+=(BMax-dis[i])/minn+1; }printf("%lld ",ans); }