Atcoder Beginner Contest 304——A-D题讲解

蒟蒻来讲题，还望大家喜。若哪有问题，大家尽可提！

Hello, 大家好哇！本初中生蒟蒻讲解一下AtCoder Beginner Contest 304这场比赛的A-D题！

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A - First Player

题目描述

Problem Statement

There are $N$ people numbered $1,2,\dots ,N$, sitting in this clockwise order around a round table.
In particular, person $1$ is sitting next to person $N$ in the clockwise direction.
For each $i=1,2,\dots ,N$, person $i$ has a name $S_i$ and an age $A_i$.
Here, no two people have the same name or the same age.
Starting from the youngest person, print the names of all $N$ people in the order of their seating positions in clockwise order.

Constraints

$2\le N\le 100$
$N$ is an integer.
$S_i$ is a string of length between $1$ and $10$, consisting of lowercase English letters.
$i\ne j⟹S_i\ne S_j$
$0\le A_i\le 10^9$
$A_i$ is an integer.
$i\ne j⟹A_i\ne A_j$

Input

The input is given from Standard Input in the following format:
$N$
$S_1$ $A_1$
$S_2$ $A_2$
$⋮$
$S_N$ $A_N$

Output

Print $N$ lines.
For each $i=1,2,\dots ,N$, the $i$-th line should contain the name of the person sitting in the $i$-th position clockwise from the youngest person.

Sample Input 1

5
alice 31
bob 41
carol 5
dave 92
ellen 65

Sample Output 1

carol
dave
ellen
alice
bob

The youngest person is person $3$. Therefore, starting from person $3$, print the names in the clockwise order of their seating positions: person $3$, person $4$, person $5$, person $1$, and person $2$.

Sample Input 2

2
takahashi 1000000000
aoki 999999999

Sample Output 2

aoki
takahashi

题目翻译

本题的意思为从年龄最低的那个人开始按顺时针方向输出人名。

思路

1. 找到年龄最低的人
2. 顺时针输出人名

代码

#include <iostream>

using namespace std;

int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    int n;
    string s[101];
    int a[101];
    
    cin >> n;
   
    int mn = 0x3f3f3f3f, pos;
    for (int i =1 ;i <= n; i ++)
    {
    	cin >> s[i] >> a[i];
    	if (a[i] < mn)
    		mn = a[i], pos = i;
    }
    
    for (int i = pos; i <= n; i ++) cout << s[i] << endl;
    for (int i = 1; i < pos; i ++) cout << s[i] << endl;
    
    return 0;
}

B - Subscribers

题目描述

Problem Statement

You are given an integer $N$.

Print an approximation of $N$ according to the following instructions.
If $N$ is less than or equal to $10^3-1$, print $N$ as it is.
If $N$ is between $10^3$ and $10^4-1$, inclusive, truncate the ones digit of $N$ and print the result.
If $N$ is between $10^4$ and $10^5-1$, inclusive, truncate the tens digit and all digits below it of $N$ and print the result.
If $N$ is between $10^5$ and $10^6-1$, inclusive, truncate the hundreds digit and all digits below it of $N$ and print the result.
If $N$ is between $10^6$ and $10^7-1$, inclusive, truncate the thousands digit and all digits below it of $N$ and print the result.
If $N$ is between $10^7$ and $10^8-1$, inclusive, truncate the ten-thousands digit and all digits below it of $N$ and print the result.
If $N$ is between $10^8$ and $10^9-1$, inclusive, truncate the hundred-thousands digit and all digits below it of $N$ and print the result.

Constraints

$N$ is an integer between $0$ and $10^9-1$, inclusive.

Input

The input is given from Standard Input in the following format:

$N$

Output

Print the answer.

Sample Input 1

20230603

Sample Output 1

20200000

$20230603$ is between $10^7$ and $10^8-1$ (inclusive).

Therefore, truncate the ten-thousands digit and all digits below it, and print $20200000$.

Sample Input 2

0

Sample Output 2

0

Sample Input 3

304

Sample Output 3

304

Sample Input 4

500600

Sample Output 4

500000

题目大意

• 如果$N$小于或等于$10^3-1$，则按原样打印$N$。

• 如果$N$在$10^3$和$10^4-1$之间(包括$10^3$和$10^4-1$)，则截断$N$的个位并打印结果。

• 如果$N$在$10^4$和$10^5-1$之间(包括10^4$和10^5-1$)，则截断$N$的十位及其以下的所有数字，并打印结果。

• 如果$N$在$10^5$和$10^6-1$之间(包括$N$)，则截断$N$的百位及其以下的所有位并打印结果。

• 如果$N$在$10^6$和$10^7-1$之间(包括 $10^7-1$)，则截断$N$的千位及其以下的所有数字并打印结果。

• 如果$N$在$10^7$和$10^8-1$之间(包括$10^8-1$)，则截断$N$的万位及其以下的所有数字并打印结果。

• 如果$N$在$10^8$和$10^9-1$之间(包括$10^9-1$)，则截断$N$的十万个数字及其以下的所有数字并打印结果。

思路

通过找规律发现，其实就是输出$N$的前三位，剩余的位全部输出0。
为了方便起见，我们可以用字符串来存储$N$

代码

#include <iostream>

using namespace std;

int main()
{
	string s;
	
	cin >> s;
	
	for (int i = 0; i < min((int)s.size(), 3); i ++)
		cout << s[i];
	for (int i = 3; i < s.size(); i ++)
		cout << 0;
}

C - Virus

题目描述

Problem Statement

There are $N$ people numbered $1,2,\dots ,N$ on a two-dimensional plane, and person $i$ is at the point represented by the coordinates $(X_i,Y_i)$.
Person $1$ has been infected with a virus. The virus spreads to people within a distance of $D$ from an infected person.
Here, the distance is defined as the Euclidean distance, that is, for two points $(a_1,a_2)$ and $(b_1,b_2)$, the distance between these two points is $\sqrt{(a_1-b_1{)}^2+(a_2-b_2{)}^2}$.
After a sufficient amount of time has passed, that is, when all people within a distance of $D$ from person $i$ are infected with the virus if person $i$ is infected, determine whether person $i$ is infected with the virus for each $i$.

Constraints

$1\le N,D\le 2000$
$-1000\le X_i,Y_i\le 1000$
$(X_i,Y_i)\ne (X_j,Y_j)$ if $i\ne j$.
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$ $D$
$X_1$ $Y_1$
$X_2$ $Y_2$
$⋮$
$X_N$ $Y_N$

Output

Print $N$ lines. The $i$-th line should contain Yes if person $i$ is infected with the virus, and No otherwise.

Sample Input 1

4 5
2 -1
3 1
8 8
0 5

Sample Output 1

Yes
Yes
No
Yes

The distance between person $1$ and person $2$ is $\sqrt{5}$, so person $2$ gets infected with the virus.

Also, the distance between person $2$ and person $4$ is $5$, so person $4$ gets infected with the virus.

Person $3$ has no one within a distance of $5$, so they will not be infected with the virus.

Sample Input 2

3 1
0 0
-1000 -1000
1000 1000

Sample Output 2

Yes
No
No

Sample Input 3

9 4
3 2
6 -1
1 6
6 5
-2 -3
5 3
2 -3
2 1
2 6

Sample Output 3

Yes
No
No
Yes
Yes
Yes
Yes
Yes
No

题目翻译

本题的意思就是说第一个人被感染上了传染病（从传染病的角度看，这是传染源 ），后来只要与他相隔的距离小于等于$D$的易感人群，都会被传染，此时这些人就会成为传播途径，再去传染其他人（帮大家复习一下生物，嘿嘿 ）

思路

这道题我们可以用bfs，首先队列中有第一个人，之后进行枚举，与他距离小于等于$D$的人放入队列，同时记录他已被传染。最后输出即可。

详情见代码~~~

代码

#include <iostream>
#include <cmath>
#include <queue>

using namespace std;

const int N = 2e3 + 10;

int x[N], y[N];
bool res[N], st[N];

int main()
{
	int n, d;
	
	cin >> n >> d;
	
	for (int i = 1; i <= n; i ++)
		cin >> x[i] >> y[i];
		
	queue<int> patient;
	
	patient.push(1), res[1] = 1;
	while (patient.size())
	{
		auto t = patient.front();
		patient.pop();
		
		if (st[t]) continue;
		st[t] = 1;
		
		for (int i = 1; i <= n; i ++)
			if (i != t && sqrt(abs(x[t] - x[i]) * abs(x[t] - x[i]) + abs(y[t] - y[i]) * abs(y[t] - y[i])) <= d)
				res[i] = 1, patient.push(i);
	}
	
	for (int i = 1; i <= n; i ++)
		cout << (res[i] ? "Yes" : "No") << endl;
}

D - A Piece of Cake

题目描述

Problem Statement

There is a rectangular cake with some strawberries on the $xy$-plane. The cake occupies the rectangular area $\{(x,y):0\le x\le W,0\le y\le H\}$.
There are $N$ strawberries on the cake, and the coordinates of the $i$-th strawberry are $(p_i,q_i)$ for $i=1,2,\dots ,N$. No two strawberries have the same coordinates.
Takahashi will cut the cake into several pieces with a knife, as follows.
First, cut the cake along $A$ different lines parallel to the $y$-axis: lines $x=a_1$, $x=a_2$, $\dots$, $x=a_A$.
Next, cut the cake along $B$ different lines parallel to the $x$-axis: lines $y=b_1$, $y=b_2$, $\dots$, $y=b_B$.
As a result, the cake will be divided into $(A+1)(B+1)$ rectangular pieces. Takahashi will choose just one of these pieces to eat. Print the minimum and maximum possible numbers of strawberries on the chosen piece.
Here, it is guaranteed that there are no strawberries along the edges of the final pieces. For a more formal description, refer to the constraints below.

Constraints

$3\le W,H\le 10^9$
$1\le N\le 2\times 10^5$
$0<p_i<W$
$0<q_i<H$
$i\ne j⟹(p_i,q_i)\ne (p_j,q_j)$
$1\le A,B\le 2\times 10^5$
$0<a_1<a_2<\cdots <a_A<W$
$0<b_1<b_2<\cdots <b_B<H$
p i ∉ { a 1 , a 2 , … , a A } p_i \not \in \lbrace a_1, a_2, \ldots, a_A \rbrace pi​∈{a1​,a2​,…,aA​}
q i ∉ { b 1 , b 2 , … , b B } q_i \not \in \lbrace b_1, b_2, \ldots, b_B \rbrace qi​∈{b1​,b2​,…,bB​}
All input values are integers.

Input

The input is given from Standard Input in the following format:

$W$ $H$
$N$
$p_1$ $q_1$
$p_2$ $q_2$
$\vdots$
$p_N$ $q_N$
$A$
$a_1$ $a_2$ $\ldots$ $a_A$
$B$
$b_1$ $b_2$ $\ldots$ $b_B$

Output

Print the minimum possible number of strawberries $m$ and the maximum possible number $M$ on the chosen piece in the following format, separated by a space.

$m$ $M$

Sample Input 1

7 6
5
6 1
3 1
4 2
1 5
6 2
2
2 5
2
3 4

Sample Output 1

0 2

There are nine pieces in total: six with zero strawberries, one with one strawberry, and two with two strawberries. Therefore, when choosing just one of these pieces to eat, the minimum possible number of strawberries on the chosen piece is $0$, and the maximum possible number is $2$.

Sample Input 2

4 4
4
1 1
3 1
3 3
1 3
1
2
1
2

Sample Output 2

1 1

Each piece has one strawberry on it.

思路

草莓会被分成如图的几块区域，如图：

我们想要统计处最多的与最少的，可以对于每一个点，找与他相邻的边并把个数加1。

之后我们枚举有点的边得组合，找最大值与最小值。其中最小值需特判，如果$有点的边得组合数<(A+1)\times (B+1)$那么就是0。

详情见代码~~~

代码

#include <iostream>
#include <map>
#include <vector>
#define x first
#define y second
#define int long long

using namespace std;

const int N = 2e5 + 10;
typedef pair<int, int> PII;

int h, w, m;
int u, v;
int A, B;
int a[N], b[N];
vector<PII> pt;
map<PII, int> area;

signed main()
{
	cin >> h >> w >> m;
	
	while (m --)
	{
		cin >> u >> v;
		pt.push_back({u, v});
	}
	
	cin >> A;
	for (int i = 1; i <= A; i ++)
		cin >> a[i];
		
	cin >> B;
	for (int j = 1; j <= B; j ++)
		cin >> b[j];
		
	for (auto c : pt)
	{
		int pa = lower_bound(a + 1, a + 1 + A, c.x) - a;
		int pb = lower_bound(b + 1, b + 1 + B, c.y) - b;
		
		area[{pa, pb}] ++;
	}
	
	int mx = 0, mn = 0x3f3f3f3f, num = 0;
	for (auto c : area)
		mn = min(mn, c.y), mx = max(mx, c.y), num ++;
		
	if (num < (A + 1) * (B +1))
		mn = 0;
		
	cout << mn << " " << mx << endl;
}

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