題目鏈接
C. Minimum Value Rectangle
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You have
n
sticks of the given lengths.
Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.
Let
S
be the area of the rectangle and
P
be the perimeter of the rectangle.
The chosen rectangle should have the value
P
2
S
minimal possible. The value is taken without any rounding.
If there are multiple answers, print any of them.
Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer
T
(
T
≥
1
) — the number of lists of sticks in the testcase.
Then
2
T
lines follow — lines
(
2
i
−
1
)
and
2
i
of them describe the
i
-th list. The first line of the pair contains a single integer
n
(
4
≤
n
≤
10
6
) — the number of sticks in the
i
-th list. The second line of the pair contains
n
integers
a
1
,
a
2
,
…
,
a
n
(
1
≤
a
j
≤
10
4
) — lengths of the sticks in the
i
-th list.
It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.
The total number of sticks in all
T
lists doesn’t exceed
10
6
in each testcase.
Output
Print
T
lines. The
i
-th line should contain the answer to the
i
-th list of the input. That is the lengths of the four sticks you choose from the
i
-th list, so that they form a rectangle and the value
P
2
S
of this rectangle is minimal possible. You can print these four lengths in arbitrary order.
If there are multiple answers, print any of them.
Example
inputCopy
3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5
outputCopy
2 7 7 2
2 2 1 1
5 5 5 5
Note
There is only one way to choose four sticks in the first list, they form a rectangle with sides
2
and
7
, its area is
2
⋅
7
14
, perimeter is
2
(
2
+
7
)
18
.
18
2
14
≈
23.143
.
The second list contains subsets of four sticks that can form rectangles with sides
(
1
,
2
)
,
(
2
,
8
)
and
(
1
,
8
)
. Their values are
6
2
2
18
,
20
2
16
25
and
18
2
8
40.5
, respectively. The minimal one of them is the rectangle
(
1
,
2
)
.
You can choose any four of the
5
given sticks from the third list, they will form a square with side
5
, which is still a rectangle with sides
(
5
,
5
)
.題意：是說給我們一些長度的火柴棍，讓我們建立一個矩形，並且矩形的面積和長度得滿足那個公式；輸出需要的火柴棍
思路：把那個公式展開一下，化簡化簡就是基本不等式，然後只要邊長儘可能的靠近就可以；
注意點：
1.給了數據大小不是白給的，數組得按照那個標準來
2.然後就是memset可能會導致超時，
3.再有就是無窮大賦值的時候，賦值給整型 的變量肯定白搭呀！
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int a[10000];//每個數據大小不超過1e4
int b[1000000];//總共有1e6個數據
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,p=0,i;
for(i=0;i<=10000;i++)a[i]=0;//一般超時可能會是memset的原因，改成這樣就沒問題了
scanf("%d",&n);
for(i=1; i<=n; i++)
{
int m;
scanf("%d",&m);
a[m]++;
if(a[m]==2||a[m]==4)//凡是出現了4次或者2次的都加進去，都可以做邊
b[++p]=m;
}
sort(b+1,b+1+p);
int x,y;
double minn=inf;//數據辣麼大，我還用整形來比較，不wa纔怪
for(i=1; i<p; i++)
{
if(b[i]==b[i+1])
{
x=b[i],y=b[i+1];//如果相等，就是出現了大於等於4次的情況
break;
}
double m=b[i+1]*1.0/b[i];//由於那個公式，化簡一下P^2/S,邊長是a,b的話，就是4*(a/b+b/a+2);
//基本不等式只有當a==b的時候等式才成立，在這個題中只有當a和b靠的很近的時候才滿足
if(m<minn){minn=m;x=b[i];y=b[i+1];}
}
printf("%d %d %d %d\n",x,x,y,y);
}
return 0;
}