2020牛客暑期多校训练营（第五场）——E Bogo Sort

题目描述

Today Tonnnny the monkey learned a new algorithm called Bogo Sort. The teacher gave Tonnnny the code of Bogo sort:

bool is_sorted(int a[], int n) {
    for (int i = 1; i < n; i++) {
        if (a[i] < a[i - 1]) {
            return false;
        }
    }
    return true;
}
void bogo_sort(int a[], int n) {
    while (!is_sorted(a, n)) {
        shuffle(a, a + n);
    }
}

The teacher said the shuffle function is to uniformly randomly permute the array a with length n, and the algorithm's expectation complexity is O(n⋅n!).However, Tonnnny is a determined boy — he doesn't like randomness at all! So Tonnnny improved Bogo Sort. He had chosen one favourite permutation p with length N , and he replaced the random shuffle with shuffle of p, so the improved algorithm, Tonnnny Sort, can solve sorting problems for length N array — at least Tonnnny thinks so.

int p[N] = {....}; // Tonnnny's favorite permutation of n
void shuffle(int a[], int n) {
    int b[n];
    for (int i = 0; i < n; i++) {
        b[i] = a[i]
    }
    for (int i = 0; i < n; i++) {
        a[i] = b[p[i]];
    }
}

void tonnnny_sort(int a[], int n) {
    assert (n == N); // Tonnnny appointed!
    while (!is_sorted(a, n)) {
        shuffle(a, a + n);
    }
}

Tonnnny was satsified with the new algorithm, and decided to let you give him a different array of length N every day to sort it with Tonnnny Sort.You are the best friend of Tonnnny. Even though you had found the algorithm is somehow wrong, you want to make Tonnnny happy as long as possible. You're given N ,p and you need to calculate the maximum number of days that Tonnnny will be happy, since after that you can't give Tonnnny an array that can be sorted with Tonnnny Sort and didn't appeared before.The answer may be very large. Tonnnny only like numbers with at most N digits, so please output answer mod $10^N$ instead.

输入描述

The first line contains one integer N (1≤ N ≤ $10^5$ ).The second line contains N integer indicating p, which forms a permutation of 1,2,⋯,N .

输出描述

The maximum number of days that Tonnnny will be happy, module $10^N$ .

输入

输入1

5
1 2 3 4 5

输入2

6
2 3 4 5 6 1

输出

输出1

输出2

题目大意

今天，Tonnnny学习了一种称为Bogo Sort的新算法。老师给了Tonnnny Bogo Sort的代码：

bool is_sorted(int a[], int n) {
    for (int i = 1; i < n; i++) {
        if (a[i] < a[i - 1]) {
            return false;
        }
    }
    return true;
}
void bogo_sort(int a[], int n) {
    while (!is_sorted(a, n)) {
        shuffle(a, a + n);
    }
}

老师说，函数shuffle 是等概率地随机排列长度为n 的数组a，这个算法的期望复杂度为O(n⋅n!)。

但是，Tonnnny是一个坚定的男孩——他一点也不喜欢随机！因此，Tonnnny改进了Bogo Sort。他选择了一个最喜欢的长度为N 的排列 p，然后用排列 p 替换随机的shuffle，因此改进的算法，Tonnnny Sort，可以解决长度为N 的数组的排序问题——至少Tonnnny如此认为。

int p[N] = {....}; // Tonnnny最喜欢的n的排列
void shuffle(int a[], int n) {
    int b[n];
    for (int i = 0; i < n; i++) {
        b[i] = a[i]
    }
    for (int i = 0; i < n; i++) {
        a[i] = b[p[i]];
    }
}

void tonnnny_sort(int a[], int n) {
    assert (n == N); // Tonnnny指定的
    while (!is_sorted(a, n)) {
        shuffle(a, a + n);
    }
}

Tonnnny对新算法感到满意，因此决定让您每天给他一个长度为N的不同数组，以使用Tonnnny Sort对其进行排序。

您是Tonnnny最好的朋友。即使您发现该算法有某种错误，也希望让Tonnnny多快乐一会。给定 N 和 p，您需要计算Tonnnny可以开心的最大天数，因为在那之后，您将不能给Tonnnny一个可以用Tonnnny Sort排序的之前没出现过的数组。

答案可能非常大。 Tonnnny只喜欢最多N 位的数字，因此请改为输出答案mod $10^N$ 。

题解

所有环长度的 LCM。由于环长度总和是 n，所以一定不会大于 n 位数，不取模即可。最长的答案也就几百位。

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int n,l,r,cn=1,an,le=1,a[maxn],b[maxn],c[maxn],d[maxn],e[maxn],g[maxn],nu[maxn];
bool f[maxn];
void co(int x)
{
	for(int i=1;i<=le;i++)nu[i]*=x;
	for(int i=1;i<le;i++)if(nu[i]>9)nu[i+1]+=nu[i]/10,nu[i]%=10;
	while(le<n&&nu[le]>9)nu[le+1]+=nu[le]/10,nu[le++]%=10;
}
int main()
{
	nu[1]=1;
	for(int i=2;i<maxn;i++)if(!d[i]){for(int j=i*2;j<maxn;j+=i)d[j]=1;e[++an]=i;}
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	{
		l=i,r=0;
		while(!f[l])f[l]=1,r=1,c[cn]++,l=a[l];
		if(r)cn++;
	}
	for(int i=0;i<cn;i++)
	{
		for(int j=1,t=0;j<an&&c[i]>1;j++)
		{
			while(c[i]%e[j]==0)t++,c[i]/=e[j];
			b[e[j]]=max(b[e[j]],t);t=0;
		}
	}
	for(int i=1;i<=an;i++)while(b[e[i]]--)co(e[i]);
	for(int i=le;i>=1;i--)printf("%d",nu[i]);
}