AtCoder Beginner Contest 297——A-E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 297这场比赛的A-E题！

今晚比前面几场要简单点，但我在B题翻了下车，第一次提交竟然WA了，做题要仔细啊。

开心的是，今晚终于进到绿名了！

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A - Double Click

原题

Problem Statement

Takahashi turned on a computer at time $0$ and clicked the mouse $N$ times. The $i$-th $(1\le i\le N)$ click was at time $T_i$.
If he consecutively clicked the mouse at time $x_1$ and time $x_2$ (where KaTeX parse error: Expected 'EOF', got '&' at position 5: x_1 &̲lt; x_2), a double click is said to be fired at time $x_2$ if and only if $x_2-x_1\le D$.
What time was a double click fired for the first time? If no double click was fired, print -1 instead.

Constraints

$1\le N\le 100$
$1\le D\le 10^9$
$1\le T_i\le 10^9(1\le i\le N)$
KaTeX parse error: Expected 'EOF', got '&' at position 5: T_i &̲lt; T_{i+1}(1 \…
All values in the input are integers.

Input

The input is given from Standard Input in the following format:
$N$ $D$
$T_1$ $T_2$ $\dots$ $T_N$

Output

If at least one double click was fired, print the time of the first such event; otherwise, print -1.

Sample Input 1

4 500
300 900 1300 1700

Sample Output 1

1300
Takahashi clicked the mouse at time $900$ and $1300$. Since $1300-900\le 500$, a double click was fired at time $1300$.
A double click had not been fired before time $1300$, so $1300$ should be printed.

Sample Input 2

5 99
100 200 300 400 500

Sample Output 2

-1
No double click was fired, so print -1.

Sample Input 3

4 500
100 600 1100 1600

Sample Output 3

600
If multiple double clicks were fired, be sure to print only the first such event.

思路

题目简单，不写了

代码

#include 
#include 
#include 
#define endl '\n'
#define psb(i) push_back(i)
#define ppb() pop_back()
#define psf(i) push_front(i)
#define ppf() pop_front()
#define ps(i) push(i)
 
using namespace std;
 
typedef pair<int, int> PII;
 
const int N = 1e2 + 10;
 
int n, d;
int a[N];
 
inline int read()
{
    int w = 1, s = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    
    return w * s;
}
 
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    cin >> n >> d;
    
    for (int i = 1; i <= n; i ++)
    	cin >> a[i];
    	
    for (int i = 2; i <= n; i ++)
    	if (a[i] - a[i - 1] <= d)
    	{
    		cout << a[i];
    		return 0;
    	}
    	
    cout << -1 << endl;
    
    return 0;
}

---

B - chess960

原题

Problem Statement

Takahashi is playing a game called Chess960. He has decided to write a code that determines if a random initial state satisfies the conditions of Chess960.

You are given a string $S$ of length eight. $S$ has exactly one K and Q, and exactly two R's, B's , and N's. Determine if $S$ satisfies all of the following conditions. Suppose that the $x$-th and $y$-th $(x < y)$ characters from the left of $S$ are B; then, $x$ and $y$ have different parities. K is between two R's. More formally, suppose that the $x$-th and $y$-th $(x < y)$ characters from the left of $S$ are R and the $z$-th is K; then $x< z < y$. #### Constraints

$S$ is a string of length $8$ that contains exactly one K and Q, and exactly two R's, B's , and N's.

Input

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

Output

Print Yes if $S$ satisfies the conditions; print No otherwise.

Sample Input 1

RNBQKBNR

Sample Output 1

Yes
The $3$-rd and $6$-th characters are B, and $3$ and $6$ have different parities.
Also, K is between the two R's. Thus, the conditions are fulfilled.

Sample Input 2

KRRBBNNQ

Sample Output 2

No
K is not between the two R's.

Sample Input 3

BRKRBQNN

Sample Output 3

No

思路

题目简单，但要注意不要理解错题意，下标奇偶性不同是字母B，K是在两个R之间

代码

#include 
#include 
#include 
#include 
#define endl '\n'
#define psb(i) push_back(i)
#define ppb() pop_back()
#define psf(i) push_front(i)
#define ppf() pop_front()
#define ps(i) push(i)
 
using namespace std;
 
typedef pair<int, int> PII;
 
string s;
vector<int> pl;
 
inline int read()
{
    int w = 1, s = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    
    return w * s;
}
 
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    cin >> s;
    
    int has= 0, k = 0;
    for (int i = 0; i < s.size(); i ++)
    {
    	if (s[i] == 'B')
    		pl.psb(i + 1);
    	if (s[i] == 'K' && has != 1)
    	{
    		cout << "No" << endl;
    		return 0;
    	}
    	if (s[i] == 'R')
    		has ++;
    }
    
    if (pl[0] % 2 != pl[1] % 2)
    	cout << "Yes" << endl;
    else
    	cout << "No" << endl;
    
    return 0;
}

---

C - PC on the Table

原题

Problem Statement

Planning to place many PCs in his room, Takahashi has decided to write a code that finds how many PCs he can place in his room.

You are given $H$ strings $S_1,S_2,\ldots,S_H$, each of length $W$, consisting of . and T. Takahashi may perform the following operation any number of times (possibly zero): Choose integers satisfying $1\leq i \leq H$ and $1 \leq j \leq W-1$ such that the $j$-th and $(j+1)$-th characters of $S_i$ are both T. Replace the $j$-th character of $S_i$ with P, and $(j+1)$-th with C. He tries to maximize the number of times he performs the operation. Find possible resulting $S_1,S_2,\ldots,S_H$. #### Constraints

$1\le H\le 100$
$2\le W\le 100$
$H$ and $W$ are integers.
$S_i$ is a string of length $W$ consisting of . and T.

Input

The input is given from Standard Input in the following format:
$H$ $W$
$S_1$
$S_2$
$⋮$
$S_H$

Output

Print a sequence of strings, $S_1,S_2,\dots ,S_H$, separated by newlines, possibly resulting from maximizing the number of times he performs the operation.
If multiple solutions exist, print any of them.

Sample Input 1

2 3
TTT
T.T

Sample Output 1

PCT
T.T
He can perform the operation at most once.
For example, an operation with $(i,j)=(1,1)$ makes $S_1$ PCT.

Sample Input 2

3 5
TTT…
.TTT.
TTTTT

Sample Output 2

PCT…
.PCT.
PCTPC

思路

找相邻的T更改为PC即可！

代码

#include 
#include 
#include 
#define endl '\n'
#define psb(i) push_back(i)
#define ppb() pop_back()
#define psf(i) push_front(i)
#define ppf() pop_front()
#define ps(i) push(i)
 
using namespace std;
 
typedef pair<int, int> PII;
 
const int N = 1e2 + 10;
 
int r, c;
char a[N][N];
 
inline int read()
{
    int w = 1, s = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    
    return w * s;
}
 
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    cin >> r >> c;
    
    for (int i = 1; i <= r; i ++)
    	for (int j = 1; j <= c; j ++)
    		cin >> a[i][j];
    		
    for (int i = 1; i <= r; i ++)
    	for (int j = 1; j < c; j ++)
    		if (a[i][j] == 'T' && a[i][j] == a[i][j + 1])
    		{
    			a[i][j] = 'P';
    			a[i][j + 1] = 'C';
    			j ++;
    		}
    
    for (int i = 1; i <= r; i ++)
    {
    	for (int j = 1; j <= c; j ++)
    		cout << a[i][j];
    	cout << endl;
    }
    return 0;
}

---

D - Count Subtractions

原题

Problem Statement

You are given positive integers $A$ and $B$.
You will repeat the following operation until $A=B$:
compare $A$ and $B$ to perform one of the following two:
if KaTeX parse error: Expected 'EOF', got '&' at position 3: A &̲gt; B, replace $A$ with $A-B$;
if KaTeX parse error: Expected 'EOF', got '&' at position 3: A &̲lt; B, replace $B$ with $B-A$.
How many times will you repeat it until $A=B$? It is guaranteed that a finite repetition makes $A=B$.

Constraints

$1\le A,B\le 10^{18}$
All values in the input are integers.

Input

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

Output

Print the answer.

Sample Input 1

3 8

Sample Output 1

4
Initially, $A=3$ and $B=8$. You repeat the operation as follows:
KaTeX parse error: Expected 'EOF', got '&' at position 2: A&̲lt;B, so replace $B$ with $B-A=5$, making $A=3$ and $B=5$.
KaTeX parse error: Expected 'EOF', got '&' at position 2: A&̲lt;B, so replace $B$ with $B-A=2$, making $A=3$ and $B=2$.
KaTeX parse error: Expected 'EOF', got '&' at position 2: A&̲gt;B, so replace $A$ with $A-B=1$, making $A=1$ and $B=2$.
KaTeX parse error: Expected 'EOF', got '&' at position 2: A&̲lt;B, so replace $B$ with $B-A=1$, making $A=1$ and $B=1$.
Thus, you repeat it four times.

Sample Input 2

1234567890 1234567890

Sample Output 2

0
Note that the input may not fit into a 32-bit integer type.

Sample Input 3

1597 987

Sample Output 3

15

题意

A和B两个数，每次操作将大数变成大数减小数的差值，直到A和B相等时停止，问一共需要操作多少次？

思路

对于A和B，因为最终A==B，所以操作过程中，谁是A，谁是B并不重要。我们不妨设A始终大于B，即把大数给A，那么会出现以下情况：
设cnt=0，用来记录操作数。
情况1：如果A=B，直接输出cnt；
情况2：如果A能整除B，那么cnt+=A/B，输出cnt；
情况3：如果A不能整除B，那么cnt+=A/B，A=A%B；
此时重新回到开始，比较新的A和B，让A是大数，B是小数，再判断以上三种情况。

本题比较简单，如果有兴趣，可以做一下AtCoder Regular Contest 159的B题，难度比这个题目高不少，但作法类似。

代码

#include 
#include 
#include 
#define int long long
#define endl '\n'
#define psb(i) push_back(i)
#define ppb() pop_back()
#define psf(i) push_front(i)
#define ppf() pop_front()
#define ps(i) push(i)
 
using namespace std;
 
typedef pair<int, int> PII;
 
int a, b;
 
inline int read()
{
    int w = 1, s = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    
    return w * s;
}
 
signed main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    cin >> a >> b;
    	
    int cnt = 0;
    while (1)
    	if (a == b)
    	{
    		cout << cnt << endl;
    		return 0;
    	}
    	else if (a < b)
    		swap(a, b);
    	else
    	{
    		if (a % b == 0)
    		{
    			cnt += a / b - 1;
    			cout << cnt << endl;
    			return 0;
    		}
    		else
    		{
    			cnt += a / b;
    			a = a % b;
    		}
    	}
    
    return 0;
}

---

E - Kth Takoyaki Set

原题

Problem Statement

In AtCoder Kingdom, $N$ kinds of takoyakis (ball-shaped Japanese food) are sold. A takoyaki of the $i$-th kind is sold for $A_i$ yen.
Takahashi will buy at least one takoyaki in total. He is allowed to buy multiple takoyakis of the same kind.
Find the $K$-th lowest price that Takahashi may pay. Here, if there are multiple sets of takoyakis that cost the same price, the price is counted only once.

Constraints

$1\le N\le 10$
$1\le K\le 2\times 10^5$
$1\le A_i\le 10^9$
All values in the input are integers.

Input

The input is given from Standard Input in the following format:
$N$ $K$
$A_1$ $A_2$ $\dots$ $A_N$

Output

Print the answer as an integer.

Sample Input 1

4 6
20 25 30 100

Sample Output 1

50
The four kinds of takoyakis sold in AtCoder Kingdom cost $20$ yen, $25$ yen, $30$ yen, and $100$ yen.
The six lowest prices that Takahashi may pay are $20$ yen, $25$ yen, $30$ yen, $40$ yen, $45$ yen, and $50$ yen. Thus, the answer is $50$.
Note that at least one takoyaki must be bought.

Sample Input 2

2 10
2 1

Sample Output 2

10
Note that a price is not counted more than once even if there are multiple sets of takoyakis costing that price.

Sample Input 3

10 200000
955277671 764071525 871653439 819642859 703677532 515827892 127889502 881462887 330802980 503797872

Sample Output 3

5705443819

题意

给定一个N个元素的整数序列a，序列中元素任意组合（同一元素可以多次）的和，判断第K小的和是多少。
如样例1：
a={20,25,30,100}
那么最小的数是20，其次依次是25、30、40（20+20）、45（20+25）、50（20+30 或者 25+25）、55（25+20）、60（30+30）……
所以K=6时，第6小的是50。

思路

这个题目可以这样理解，以样例1为例，最小的数是直接可以知道的，即20，那么从20往后，每次都是用当前数加一遍序列里的所有数，然后把当前数抛掉，再找下一个最小数，这个最小数再把序列里的所有数加一遍，再重复，直到第K个数，用个数轴来表示更清晰一点：

---

代码

#include 
#include 
#include 
#include 
#define endl '\n'
#define int long long
#define psb(i) push_back(i)
#define ppb() pop_back()
#define psf(i) push_front(i)
#define ppf() pop_front()
#define ps(i) push(i)

using namespace std;

typedef pair<int, int> PII;

const int N = 15;

int n, k;
int a[N];
set<int> s;

inline int read()
{
    int w = 1, s = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    
    return w * s;
}

signed main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    
    cin >> n >> k;
    
    for (int i = 1; i <= n; i ++)
    	cin >> a[i];
    
	s.insert(0);
    	
    int cnt = 0;
    while (cnt < k)
    {
    	for (int i = 1; i <= n; i ++)
    		s.insert(*s.cbegin() + a[i]);
    		
    	s.erase(*s.cbegin());
    	cnt ++;
    }
	
	cout << *s.cbegin() << endl;
    
    return 0;
}

---

今天就到这里了！

大家有什么问题尽管提，我都会尽力回答的！

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