# UVA12558 埃及分数

#include<iostream>
#include<cstdio>
#include<set>
#include<memory.h>
using namespace std;
#define int long long
#define max(a,b) ((a)>(b)?(a):(b))
#define N 1000010
set<int> S;
int T,a,b,n;
int d;
int r[N],ans[N],v[N];
bool op;
int gcd(int x,int y){
return y?gcd(y,x%y):x;
}
inline bool compare(int x){
for(int i=x;i>=1;i--)
if(r[i]!=ans[i])
return ans[i]==-1 || r[i]<ans[i];
return 0;
}
void dfs(int k,int t,int p,int q){
if(k==d){
if(p!=1 || S.count(q))
return ;
r[k]=q;
if(compare(d))
for(int i=1;i<=d;i++)
ans[i]=r[i];
op=1;
return ;
}
t=max(t,(q-1)/p+1);
for(;;t++){
if(q*(d-k+1)<=t*p)
break;
if(S.count(t))
continue;
r[k]=t;
int np=p*t-q,nq=q*t,nr=gcd(np,nq);
dfs(k+1,t+1,np/nr,nq/nr);
}
}
signed main(){
scanf("%d",&T);
for(int i=1;i<=T;i++){
op=0;
S.clear();
scanf("%lld%lld%lld",&a,&b,&n);
printf("Case %lld: %lld/%lld=",i,a,b);
int x;
for(int j=1;j<=n;j++){
scanf("%lld",&x);
S.insert(x);
}
for(d=1;;d++){
memset(ans,-1,sizeof(ans));
dfs(1,0,a,b);
if(op)
break;
}
for(int j=1;j<d;j++)
printf("1/%lld+",ans[j]);
printf("1/%lld\n",ans[d]);
}
return 0;
}
posted @ 2019-03-22 09:22  Pelom  阅读(...)  评论(... 编辑 收藏