BZOJ 2118 墨墨的不等式 数论 + 最短路 + 计数

#include<bits/stdc++.h>
#define LL long long
const LL INF = 50000000000000000ll; 
#define maxn 500010
using namespace std;
LL l, r;
LL ans = 0;
int n, a[maxn];
LL mi = INF, d[maxn];
LL solve(LL x){
    if(x > r) return 0;
    else if(x >= l){
        return ((r - x)/mi + 1ll);
    }
    else{
        return ((r - x)/mi - (l - 1 - x)/mi);
    }
}
void spfa(){
    bool vis[maxn];
    memset(vis, false, sizeof(vis));
    for(int i = 0; i < mi; i ++){
        d[i] = INF;
    }
    d[0] = 0;
    vis[0] = true;
    queue<int> Q;
    Q.push(0);
    int p, now;
    while(!Q.empty()){
        p = Q.front();
        Q.pop();
        vis[p] = false;
        for(int i = 1; i <= n; i ++){
            if(a[i] == mi) continue;
            now = (p + a[i]) % mi;
            if(d[now] > d[p] + a[i]){
                d[now] = d[p] + a[i];
                if(!vis[now]){
                    vis[now] = true;
                    Q.push(now);
                }
            }
        }
    }
}
int main(){
    cin >> n >> l >> r; 
    for(int i = 1; i <= n; i ++){
        scanf("%d",&a[i]);
        if(a[i] == 0){
            i --, n --;
        }
        else
        mi = min(mi, (LL)a[i]);
    }
    if(mi == 1){
        cout << (r - l + 1) << endl;
        return 0;
    }
    spfa();
    for(int i = 0; i < mi; i ++){
        if(d[i] == INF) continue;
        ans += solve((LL)d[i]);
    }    cout << ans;
    return 0;
}