BZOJ2118: 墨墨的等式

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2118

题解：考虑模某个数意义下x是否能达到。用最小的数作为模数，那么只要建图跑最短路就可以统计答案了。

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000000ll
#define maxn 500000+5
#define maxm 8000000+5
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
#define mod 1000000007
#define lch t[k].l,l,mid
#define rch t[k].r,mid+1,r
using namespace std;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
int n,head[maxn],mx=1000000,tot,a[maxn];
ll l,r,d[maxn];
bool v[maxn];
queue<int>q;
struct edge{int go,next,w;}e[maxm];
inline void add(int x,int y,int w)
{
    e[++tot]=(edge){y,head[x],w};head[x]=tot;
}
inline ll work(ll x,ll y,ll z)
{
    return x<y?0:(x-y)/z+1;
}
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=read();l=read();r=read();
    for1(i,n)a[i]=read(),mx=min(mx,a[i]);
    for1(i,n)for0(j,mx-1)add(j,(j+a[i])%mx,a[i]);
    for0(i,mx-1)d[i]=inf;
    d[0]=0;q.push(0);
    while(!q.empty())
    {
        int x=q.front();q.pop();v[x]=0;
        for4(i,x)if(d[x]+e[i].w<d[y])
        {
            d[y]=d[x]+(ll)e[i].w;
            if(!v[y]){v[y]=1;q.push(y);}
        }
    }
    ll ans=0;
    for0(i,mx-1)ans+=work(r,d[i],mx)-work(l-1,d[i],mx);
    cout<<ans<<endl;
    return 0;
}