【Uva 12558】 Egyptian Fractions (HARD version) （迭代加深搜，IDA*）

1 #include<cstdio>
2 #include<cstdlib>
3 #include<cstring>
4 #include<iostream>
5 #include<algorithm>
6 #include<queue>
7 #include<cmath>
8 using namespace std;
9 #define LL long long
10 #define Maxn 1100
11
12 LL a,b;
13 LL maxd,ans[Maxn],v[Maxn];
14
15 bool qq[1010];
16
17 LL mymax(LL x,LL y) {return x>y?x:y;}
18
19 bool better(LL d)
20 {
21     for(LL i=d;i>=1;i--) if(v[i]!=ans[i])
22     {
23         return ans[i]==-1||v[i]<ans[i];
24     }
25     return 0;
26 }
27
28 LL get_first(LL a,LL b)
29 {
30     for(LL i=1;;i++)
31     {
32         if(b<=a*i) return i;
33     }
34 }
35
36 LL gcd(LL a,LL b)
37 {
38     if(b==0) return a;
39     return gcd(b,a%b);
40 }
41
42 bool dfs(LL d,LL from,LL aa,LL bb)
43 {
44     if(d==maxd)
45     {
46         if(bb%aa) return 0;
47         v[d]=bb/aa;
48         if(v[d]<=1000&&!qq[v[d]]) return 0;
49         if(better(d)) memcpy(ans,v,sizeof(ans));
50         return 1;
51     }
52     bool ok=0;
53     from=mymax(from,get_first(aa,bb));
54     for(LL i=from;;i++)
55     {
56         if(i<=1000&&!qq[i]) continue;
57         if(bb*(maxd+1-d)<=i*aa) break;
58         v[d]=i;
59         LL b2=bb*i;
60         LL a2=aa*i-bb;
61         LL g=gcd(a2,b2);
62         if(dfs(d+1,i+1,a2/g,b2/g)) ok=1;
63     }
64     return ok;
65 }
66
67 int main()
68 {
69     LL T,kase=0;
70     scanf("%lld",&T);
71     while(T--)
72     {
73         scanf("%lld%lld",&a,&b);
74         memset(qq,1,sizeof(qq));
75         LL k;
76         scanf("%lld",&k);
77         for(LL i=1;i<=k;i++)
78         {
79             LL x;
80             scanf("%lld",&x);
81             qq[x]=0;
82         }
83         for(maxd=1;;maxd++)
84         {
85             memset(ans,-1,sizeof(ans));
86             if(dfs(1,get_first(a,b),a,b)) break;
87         }
88         printf("Case %lld: %lld/%lld=",++kase,a,b);
89         printf("1/%lld",ans[1]);
90         for(LL i=2;i<=maxd;i++) printf("+1/%lld",ans[i]);
91         printf("\n");
92         /*printf("%d\n",maxd);
93         for(LL i=1;i<=maxd;i++) printf("%d\n",ans[i]);*/
94     }
95     return 0;
96 }