BZOJ2118: 墨墨的等式

BZOJ2118: 墨墨的等式

题解：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define D(x) cout<<#x<<" = "<<x<<"  "
#define E cout<<endl
using namespace std;
typedef long long ll;
const ll N = 1e6+5;

ll n,p,low,high,a
,d
; bool vis
;

ll que(ll x){
ll res=0;
for(ll i=0;i<p;i++) if(d[i]<=x){
res+=(x-d[i])/p+1;
}
return res;
}

void spfa(){
memset(d,0x3f,sizeof(d));
queue<int> q; q.push(0); d[0]=0;
while(!q.empty()){
ll x=q.front(); q.pop(); vis[x]=false;
for(ll i=1;i<=n;i++){
ll v=(x+
4000
a[i])%p;
if(d[v]>d[x]+a[i]){
d[v]=d[x]+a[i];
if(!vis[v]){
vis[v]=true;
q.push(v);
}
}
}
}
}

int main(){
cin>>n>>low>>high; p=1e6;
for(ll i=1;i<=n;i++){
cin>>a[i];
if(!a[i]){ i--;n--;continue; }
p=min(p,a[i]);
}
spfa(); //for(ll i=0;i<p;i++)D(d[i]); E;
cout<<que(high)-que(low-1)<<endl;
}