[国家集训队]墨墨的等式

传送门

前置芝士：同余最短路

这题其实就是例题的强化版，还是以最小的a[i]为除数做与例题相同的建模方法即可。

#include <bits/stdc++.h>
using namespace std;
const long long MAXA = 5e5 + 10; 
struct node{
    long long pre, to, val;
}edge[MAXA * 20];
long long head[MAXA], tot;
long long n, bmi, bmx;
long long a[20];
long long dis[MAXA], vis[MAXA];
queue<long long> q;
void add(long long u, long long v, long long l) {
    edge[++tot] = node{head[u], v, l};
    head[u] = tot;
}
void spfa() {
    memset(dis, 0x3f, sizeof(dis));
    dis[0] = 0;
    vis[0] = 1;
    q.push(0);
    while (!q.empty()) {
        long long x = q.front(); q.pop();
        for (long long i = head[x]; i; i = edge[i].pre) {
            long long y = edge[i].to;
            if (dis[y] > dis[x] + edge[i].val) {
                dis[y] = dis[x] + edge[i].val;
                if (!vis[y]) {
                    vis[y] = 1;
                    q.push(y);
                }
            }
        }
        vis[x] = 0;
    }
}
long long solve(long long x) {
    long long ret = 0;
    for (long long i = 0; i < a[1]; i++) {
        if (dis[i] <= x) {
            ret += (x - dis[i]) / a[1] + 1;
        }
    }
    return ret;
}
int main() {
    cin >> n >> bmi >> bmx;
    for (long long i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for (long long i = 0; i < a[1]; i++) {
        for (long long j = 2; j <= n; j++) {
            add(i, (i + a[j]) % a[1], a[j]);
        }
    }
    spfa();
    cout << solve(bmx) - solve(bmi - 1);
    return 0;
}