1 /*
 2     枚举一个数 a[i]
 3     除以黄金比 得到 t=a[i]/gold
 4     找到 与 t相差最小的数
 5 */
 6 #include<cstdio>
 7 #include<iostream>
 8 #include<algorithm>
 9 #define gold 0.6180339887498949
10 #define MAXN 30010
11
12 using namespace std;
13
14 int n,a[MAXN],Lx,Ry;
15
16 double minn=1000000.0;
17
18 inline void read(int&x) {
19     x=0;int f=1;char c=getchar();
20     while(c>'9'||c<'0') {if(c=='-') f=-1;c=getchar();}
21     while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-48;c=getchar();}
22     x=x*f;
23 }
24
25 inline bool cmp(int a,int b) {
26     return a<b;
27 }
28
29 inline double abs(double x) {
30     if(x<0) return -x;
31     return x;
32 }
33
34 int main() {
35     read(n);
36     for(int i=1;i<=n;i++) read(a[i]);
37     sort(a+1,a+1+n,cmp);
38     for(int i=1;i<=n;i++) {
39         if(a[i]==0) continue;
40         double p=a[i]/gold;
41         int l=i,r=n,mid,k;
42         while(l<=r) {
43             mid=(l+r)>>1;
44             if(a[mid]>p) r=mid-1,k=r;
45             else l=mid+1;
46         }
47         if(a[k]==0) continue;
48         double temp=abs((double)a[i]/a[k]-gold);
49         if(temp<minn) {
50             minn=temp;
51             Lx=a[i];
52             Ry=a[k];
53         }
54     }
55     printf("%d\n%d\n",Lx,Ry);
56     return 0;
57 } 

 1 /*
 2     枚举两个数
 3     计算黄金比
 4     取最优值
 5 */
 6 #include<cstdio>
 7 #include<iostream>
 8 #include<algorithm>
 9 #define gold 0.6180339887498949
10 #define MAXN 30010
11
12 using namespace std;
13
14 int n,a[MAXN],Lx,Ry;
15
16 double minn=1000000.0;
17
18 inline void read(int&x) {
19     x=0;int f=1;char c=getchar();
20     while(c>'9'||c<'0') {if(c=='-') f=-1;c=getchar();}
21     while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-48;c=getchar();}
22     x=x*f;
23 }
24
25 inline bool cmp(int a,int b) {
26     return a<b;
27 }
28
29 inline double abs(double x) {
30     if(x<0) return -x;
31     return x;
32 }
33
34 inline bool judge(int x,int y) {
35     double GOLD=(double)a[x]/(double)a[y];
36     if(GOLD<gold) {
37         if(gold-GOLD<minn) {
38             minn=gold-GOLD;
39             Lx=a[x];
40             Ry=a[y];
41         }
42         return 1;
43     }
44     if(GOLD>gold) {
45         if(GOLD-gold<minn) {
46             minn=GOLD-gold;
47             Lx=a[x];
48             Ry=a[y];
49         }
50         return 0;
51     }
52 }
53
54 int main() {
55     read(n);
56     for(int i=1;i<=n;i++) read(a[i]);
57     sort(a+1,a+1+n,cmp);
58     for(int i=1;i<=n;i++) {
59         if(a[i]==0) continue;
60         double p=a[i]/gold;
61         int l=i,r=n,mid,k;
62         while(l<=r) {
63             mid=(l+r)>>1;
64             if(judge(i,mid)) r=mid-1;
65             else l=mid+1;
66         }
67     }
68     printf("%d\n%d\n",Lx,Ry);
69     return 0;
70 }