AtCoder Beginner Contest 296——D-E

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 296这场比赛的D-E题！

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D - M<=ab

原题

Problem Statement

You are given positive integers $N$ and $M$.

Find the smallest positive integer $X$ that satisfies both of the conditions below, or print $-1$ if there is no such integer.
$X$ can be represented as the product of two integers $a$ and $b$ between $1$ and $N$, inclusive. Here, $a$ and $b$ may be the same.
$X$ is at least $M$.

Constraints

$1\le N\le 10^{12}$
$1\le M\le 10^{12}$
$N$ and $M$ are integers.

Input

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

Output

Print the smallest positive integer $X$ that satisfies both of the conditions, or $-1$ if there is no such integer.

Sample Input 1

5 7

Sample Output 1

8
First, $7$ cannot be represented as the product of two integers between $1$ and $5$.

Second, $8$ can be represented as the product of two integers between $1$ and $5$, such as $8=2\times 4$.
Thus, you should print $8$.

Sample Input 2

2 5

Sample Output 2

-1
Since $1\times 1=1$, $1\times 2=2$, $2\times 1=2$, and $2\times 2=4$, only $1$, $2$, and $4$ can be represented as the product of two integers between $1$ and $2$,
so no number greater than or equal to $5$ can be represented as the product of two such integers.

Thus, you should print $-1$.

Sample Input 3

100000 10000000000

Sample Output 3

10000000000
For $a=b=100000$ $(=10^5)$, the product of $a$ and $b$ is $10000000000$ $(=10^{10})$, which is the answer.

Note that the answer may not fit into a $32$-bit integer type.

---

思路

首先，我们可以假定$a\le b$，这个是跟结果没有什么关系的！（显而易见）

其次，
$\because a\le b$
$\therefore ab\ge a^2$
$\because ab\ge M,ab离M最近$
$\therefore ab\ge M\ge a^2$
$\therefore a^2\le M$
$\therefore a\le \lceil \sqrt{M}\rceil$

所以，要想使ab离M最近，则$b=\lceil \frac{M}{a}\rceil$， $1\le a,b\le n$, $ab\ge M$

满足以上所有条件的a,b即为答案，求出最小的ab就可以了！

---

代码

#include 
#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;

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);
    
    int n, m;
    
    cin >> n >> m;
    
    int ans = 2e18;
    for (int a = 1; a <= ceil(sqrt(m)); a ++)
    {
    	int b = ceil(m * 1.0 / a);
    	if (b > n || a > n) continue;
    	if (a * b >= m && a * b < ans) ans = a * b;
    }
    
    if (ans == 2e18) cout << -1 << endl;
    else cout << ans << endl;
    
    return 0;
}

---

E题

原题

Problem Statement

You are given a sequence of $N$ numbers: $A=(A_1,A_2,\dots ,A_N)$. Here, each $A_i$ $(1\le i\le N)$ satisfies $1\le A_i\le N$.
Takahashi and Aoki will play $N$ rounds of a game. For each $i=1,2,\dots ,N$, the $i$-th game will be played as follows.

1. Aoki specifies a positive integer $K_i$. After knowing $K_i$ Aoki has specified, Takahashi chooses an integer $S_i$ between $1$ and $N$, inclusive, and writes it on a blackboard. Repeat the following $K_i$ times. Replace the integer $x$ written on the blackboard with $A_x$.

If $i$ is written on the blackboard after the $K_i$ iterations, Takahashi wins the $i$-th round; otherwise, Aoki wins.
Here, $K_i$ and $S_i$ can be chosen independently for each $i=1,2,\ldots,N$. Find the number of rounds Takahashi wins if both players play optimally to win. #### Constraints

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

Input

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

Output

Find the number of rounds Takahashi wins if both players play optimally to win.

Sample Input 1

3
2 2 3

Sample Output 1

2
In the first round, if Aoki specifies $K_1=2$, Takahashi cannot win whichever option he chooses for $S_1$: $1$, $2$, or $3$.
For example, if Takahashi writes $S_1=1$ on the initial blackboard, the two operations will change this number as follows: $1\to 2(=A_1)$, $2\to 2(=A_2)$. The final number on the blackboard will be $2(\ne 1)$, so Aoki wins.
On the other hand, in the second and third rounds, Takahashi can win by writing $2$ and $3$, respectively, on the initial blackboard, whatever value Aoki specifies as $K_i$.
Therefore, if both players play optimally to win, Takashi wins two rounds: the second and the third. Thus, you should print $2$.

Sample Input 2

2
2 1

Sample Output 2

2
In the first round, Takahashi can win by writing $2$ on the initial blackboard if $K_1$ specified by Aoki is odd, and $1$ if it is even.
Similarly, there is a way for Takahashi to win the second round. Thus, Takahashi can win both rounds: the answer is $2$.

---

题目大意

有一个N位的序列，$A_1,A_2\dots \dots A_N$
Aoki指定一个$K_i$
然后 Takahashi根据Aoki指定的$k_i$选择一个$s_i$,
最后，用$A_x$来替换$x$,一共替换$k_i$次
如果最后在黑板上的$A_x$正好是$k_i$的$i$，那么$Takahashi$胜，否则$Aoki$胜。

以样例1为例：

N=3
A={2,2,3}

情况1 ：
Aoki选择$K_1=2$，此时轮到 Takahashi选择$s_1$,那么$s_1$可以等1,2或者3.
当$s_1=1$时，把1写在黑板上，$x=1$，接下来$k_1$轮替换，第一轮，用$A_x$替换$x$，即$A_1$替换1，所以黑板上变成$A_1$即2，第二轮用A2替换2,所以黑板上变成2。最后，因为2不等于$k_1$中的1,所以Aoki胜。
同理，当s1=2时，最终也是2不等于1 ，也是Aoki胜。
当$S_1=3$时，最终是3不等于1,也是Aoki胜。可以得出结论，对于情况1,Takahashi是不可能胜的，此题目中双方都是最优手，所以只能是Aoki胜。

情况2：
Aoki选择K2 = 2,此时轮到 Takahashi选择s2,那么s2可以等1,2或者3.当Takahashi选择S2=1和2时，Takahashi都会胜，所以在这一情况下，Takahashi一定会选择他胜的方案，最终Takahashi胜。

情况3：
Aoki选择K3 = 3,此时轮到 Takahashi选择s3,那么s3可以等1,2或者3.当Takahashi选择S3=3时，Takahashi会胜，所以在这一情况下，Takahashi一定会选择他胜的方案，最终Takahashi胜。
因为情况2和情况3能胜，所以输出为2.

---

思路

基于以上分析，本题就是在找Aoki从$a_1,a_2\dots \dots a_n$选择时，有几个选择Takahashi方案能胜。

当选中Ki后，对于Takahashi来说，其实就是从1到N中选择一个数，如何能够快速到$A_x$使得$A_x$的值为$i$.可以考虑一个有向图。
如样例1：
$A=2,2,3$ => $1->2，2->2,3->3$
即：下标指向在数组中的数
当$k_1=2$时，此时需要找到如何从1-3中数中选一个数做为起点，能够快速到达一个x,使得ax=1，且这个快速到达经过的结点不能超过2+1=3个，就是所谓的只能经过k1轮。因为1-3中所有数都无法到达1个x,使得ax=1所以Takahashi不可能胜。
当$k_2=2$时，因为$2->2$,此时$a_2=2$,所以1轮就可以到$1\le k_2$,所以Takahashi可以选2
$cnt++$
当$k_3=3$时，因为$3->3$,此时$a_3=3$ ,所以1轮就可以到$1\le k_3$,所以Takahashi可以选3
$cnt++$
输出$cnt$

样例2：
$A=2,1$ 即$1->2;2->1$;
当$k_1=2$时，因为$2->1$，且1轮符合$1<2$所以cnt++
当$k_2=1$时，因为$1->2$，且1轮符合$1\le 1$,所以cnt++
输出$cnt$.

可以看出，对于此题来说，其实就是判断有向图中环上的点，只要是在环上的点，Takahashi就一定能胜，只要不是在环上的点，Takahashi就一定胜不了。所以，计算在环上的点的个数就可以AC。

判断环的方法有很多种，这里采用拓扑排序的方法，更多请移步于判断图中存在闭环的常用方法

---

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;

const int N = 2e5 + 10;

int n;
int a[N];
vector<int> g[N];
int din[N];
queue<int> q;

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;
    
    for (int i = 1; i <= n; i ++)
    	cin >> a[i], g[i].psb(a[i]), din[a[i]] ++;
    
    int res = n;
    for (int i = 1; i <= n; i ++)
    	if (!din[i])
    	{
    		res --;
    		q.ps(i);
    	}
    
    while (q.size())
    {
    	int t = q.front();
    	q.pop();
    	
    	for (auto c : g[t])
    	{
    		din[c] --;
    		if (!din[c])
    		{
    			res --;
    			q.ps(c);
    		}
    	}
    }
    
    cout << res << endl;
    
    return 0;
}