A题

题目链接

Problem Statement

There are N points in a two-dimensional plane. The coordinates of the i-th point are (xi​,yi​).

Find the maximum length of a segment connecting two of these points.

Constraints

• 2≤N≤100
• −1000≤xi​,yi​≤1000
• (xi​,yi​) ≠ (xj​,yj​) (i ≠ j)
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

N
x1​ y1​
x2​ y2​
  ⋮
xN​ yN​

Output

Print the maximum length of a segment connecting two of the points.

Your answer will be considered correct when the absolute or relative error from the judge's answer is at most .

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Sample Input 1

3
0 0
0 1
1 1

Sample Output 1

1.4142135624

For the 1-st and 3-rd points, the length of the segment connecting them is =1.41421356237…, which is the maximum length.

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Sample Input 2

5
315 271
-2 -621
-205 -511
-952 482
165 463

Sample Output 2

1455.7159750446

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题解：

        这道题数据量很小，暴力判断就行，注意答案的精度。

AC代码：

#include <bits/stdc++.h>
using namespace std;

int main()
{
	int n;
	scanf("%d",&n);
	int x[105],y[105];
	memset(x,0,sizeof(x));
	memset(y,0,sizeof(y));
	for (int i=0;i<n;i++)
		scanf("%d%d",&x[i],&y[i]);
	double ans=0.0;
	for (int i=0;i<n;i++)
	{
		for (int j=i+1;j<n;j++)
			ans=max(ans,sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])));
	}
	printf("%.8lf\n",ans);
	return 0;
}

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B题

题目链接

Problem Statement

Among the positive integers that consist of 0's and 2's when written in base 10, find the K-th smallest integer.

Constraints

• K is an integer between 1 and  (inclusive).

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Input

Input is given from Standard Input in the following format:

K

Output

Print the answer as an integer.
Here, the exact value must be printed as an integer, even if it is big. Exponential notations such as 2.34e+22, for example, or unnecessary leading zeros such as 0523 are not allowed.

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Sample Input 1

3

Sample Output 1

22

The positive integers that consist of 0's and 2's when written in base 10 are 2,20,22,… in ascending order.
The (K=) 3-rd of them, which is 22, should be printed.

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Sample Input 2

11

Sample Output 2

2022

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Sample Input 3

923423423420220108

Sample Output 3

220022020000202020002022022000002020002222002200002022002200

Note that the exact value of the answer must be printed as an integer, even if it is big.

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题解：

        看到最后一个样例，就知道最少应该拿string做（或者边处理边输出结果），实际上，把所有的‘2’变成‘1’之后，就是1~n的二进制表示法，所以这道题实际上就是要把K转换成二进制数，然后所有的‘1’输出为‘2’即可。

        有两种普遍的写法：第一种是用递归方式写，好处是输出的顺序与答案是一样的；第二种是非递归的形式，本人采用第二种方式，用一个string容器存储每一步的结果，最后将这个字符串倒过来就是最后的答案了。

AC代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
	ll k;
	cin>>k;
	string s;
	while (k>0)
	{
		if (k%2==0)
			s+="0";
		else
			s+="2";
		k/=2;
	}
	reverse(s.begin(),s.end());
	cout<<s<<endl;
	return 0;
}

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ps：本人实力较菜，只能做出这些简单题，后面的题如果有哪位大神可以解决可以私信我，如果这里面有讲的不清楚或者有错误的，望及时指出！