12558 - Egyptian Fractions (HARD version)

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#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;
}

12558 - Egyptian Fractions (HARD version)

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原文地址：http://blog.csdn.net/a197p/article/details/45318535