bzoj2118

最短路

很早以前做的了

数据范围太大，不能直接算

mn=min(a[i])

算出d[i]表示sum%mn=i最小能构成的数，这个用最短路就行了，然后计算d[i],d[i]+mn的个数统计答案

#include<bits/stdc++.h>
using namespace std;
const int N = 6e6 + 5;
int n;
long long l, r, mn = 0x3f3f3f3f;
long long a[20], d[N];
int main()
{
    scanf("%d%lld%lld", &n, &l, &r);
    for(int i = 1; i <= n; ++i) 
    {
        scanf("%d", &a[i]);
        mn = min(mn, a[i]);
    }
    priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > q;
    memset(d, 0x3f3f, sizeof(d));
    d[0] = 0;
    q.push({0, 0});
    while(!q.empty())
    {
        pair<int, int> o = q.top();
        q.pop();
        int u = o.second;
        if(d[u] < o.first) continue;
        for(int i = 1; i <= n; ++i) 
        {
            int v = (u + a[i]) % mn;
            if(d[v] <= d[u] + a[i]) continue;
            d[v] = d[u] + a[i];
            q.push({d[v], v});
        }
    }
    --l;
    long long ans = 0;
    for(int i = 0; i < mn; ++i) 
    {
        if(d[i] <= r) ans += (r - d[i]) / mn + 1;
        if(d[i] <= l) ans -= (l - d[i]) / mn + 1; 
    }
    printf("%lld
", ans);
    return 0;
}