12558 - Egyptian Fractions (HARD version)

来源：互联网发布：python iteritems 3.0 编辑：程序博客网时间：2024/05/20 05:59
#include<cstdio>#include<cstring>#include<algorithm>#include<set>using namespace std;typedef long long LL;const int maxn=10010;int maxd,t,tt;set<LL> sk;LL ans[maxn],v[maxn];LL gcd(LL a,LL b){    return b?gcd(b, a%b):a;}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 a,LL b){    if(d==maxd){        if(b%a) return false;        v[d]=b/a;        if(sk.count(b/a)) return false;        if(better(d)) memcpy(ans,v,sizeof(LL)*(d+1));        return true;    }    bool ok =false;    for(LL i=max(from,get_first(a,b)); ; ++i){        if(b*(maxd+1-d)<=i*a) break;        if(sk.count(i)) continue;        v[d]=i;        LL b2=b*i;        LL a2=a*i-b;        LL g=gcd(a2,b2);        if(dfs(d+1,i+1,a2/g,b2/g)) ok=true;    }    return ok;}int main(){    scanf("%d",&t);    while(t--)    {        LL a,b,k,sk0;        sk.clear();        scanf("%lld%lld%lld",&a,&b,&k);        while(k--) scanf("%lld",&sk0),sk.insert(sk0);        for(maxd=0;;++maxd)        {            memset(ans,-1,sizeof(ans));            if(dfs(0,get_first(a,b),a,b)) break;        }        printf("Case %d: %lld/%lld=",++tt,a,b);        for(int i=0;i<=maxd;++i)        {            if(i) printf("+");            printf("1/%lld",ans[i]);        }        printf("\n");    }    return 0;}
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