UVA-12558 Egyptian Fractions (HARD version) （IDA* 或 迭代加深搜索）

题目大意：经典的埃及分数问题。

代码如下：

# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
using namespace std;
# define LL long long

int num[5],a,b,k;
LL ans[10000],v[10000];

LL gcd(LL a,LL b)
{
    return (b==0)?a:gcd(b,a%b);
}
int get_first(LL a,LL b)
{
    return b/a+1;
}
bool judge(int val)
{
    for(int i=0;i<k;++i)
        if(val==num[i])
            return false;
    return true;
}
bool better(int d)
{
    for(int i=d;i>=0;--i)
        if(ans[i]!=v[i])
            return ans[i]==-1||v[i]<ans[i];
    return false;
}
bool dfs(int cur,int maxd,int from,LL aa,LL bb)
{
    if(cur==maxd){
        if(bb%aa)
            return false;
        v[cur]=bb/aa;
        if(!judge(v[cur]))
            return false;
        if(better(cur)){
            for(int i=0;i<=cur;++i)
                ans[i]=v[i];
        }
        return true;
    }
    bool ok=false;
    from=max(from,get_first(aa,bb));
    for(int i=from;;++i){
        while(!judge(i))
            ++i;
        if(bb*(maxd-cur+1)<=i*aa)
            break;
        v[cur]=i;
        LL b2=bb*i;
        LL a2=aa*i-bb;
        LL g=gcd(a2,b2);
        if(dfs(cur+1,maxd,i+1,a2/g,b2/g))
            ok=true;
    }
    return ok;
}
void solve()
{
    for(int maxd=1;;++maxd){
        memset(ans,-1,sizeof(ans));
        if(dfs(0,maxd,get_first(a,b),a,b)){
            printf("%d/%d=",a,b);
            for(int i=0;i<=maxd;++i)
                printf("1/%lld%c",ans[i],(i==maxd)?'
':'+');
            break;
        }
    }
}
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&a,&b,&k);
        for(int i=0;i<k;++i)
            scanf("%d",num+i);

        printf("Case %d: ",++cas);
        solve();
    }
    return 0;
}