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

来源：互联网发布：国际快递知乎编辑：程序博客网时间：2024/05/20 10:13

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;}

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