UVa 12558 - Egyptian Fractions (HARD version) [IDA*]

简单版和加强版的区别就是数据范围扩大到了long long，并且加了限制条件。用IDA*算法即可解决。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
LL maxd, ans[10240], v[10240];
set<LL> ban;

LL gcd(LL a, LL b){
    return b ? gcd(b, a % b) : a;
}
inline LL get_first(LL a, LL b){
    return b/a + 1;
}
bool better(int d){
    for(int i = d; i >= 0; --i)
        if(v[i] != ans[i])
            return ans[i]==-1 || v[i]<ans[i];
    return false;
}
bool DFS(int d, LL from, LL aa, LL bb)
{
    if(d == maxd){
        if(bb % aa || ban.count(bb/aa)) return false;
        v[d] = bb / aa;
        if(better(d))
            memcpy(ans, v, sizeof(LL)*(d+1));
        return true;
    }
    bool ok = false;
    from = max(from, get_first(aa, bb));
    for(LL i = from; ; ++i){
        if(ban.count(i)) continue;
        if(bb * (maxd+1-d) <= i * aa) break;
        v[d] = i;
        LL b2 = bb * i, a2 = aa * i - bb;
        LL g = gcd(a2, b2);
        if(DFS(d+1, i+1, a2/g, b2/g)) ok = true;
    }
    return ok;
}
int main()
{
    int T; cin >> T;
    for(int t = 1; t <= T; ++t){
        int a, b, n;
        cin >> a >> b >> n;
        ban.clear();
        while(n--){
            int x; cin >> x;
            ban.insert(x);
        }
        for(maxd = 1; ; ++maxd){
            memset(ans, -1, sizeof(ans));
            if(DFS(0, get_first(a, b), a, b))
                break;
        }
        printf("Case %d: %d/%d=1/%lld", t, a, b, ans[0]);
        for(int i = 1; i <= maxd; i++)
            printf("+1/%lld", ans[i]);
        printf("\n");
    }
    return 0;
}