poj 2369 Permutations（置换）

Permutations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2926   Accepted: 1572 Description We remind that the permutation of some final set is a one-to-one mapping of the set onto itself. Less formally, that is a way to reorder elements of the set. For example, one can define a permutation of the set {1,2,3,4,5} as follows:    This record defines a permutation P as follows: P(1) = 4, P(2) = 1, P(3) = 5, etc.  What is the value of the expression P(P(1))? It’s clear, that P(P(1)) = P(4) = 2. And P(P(3)) = P(5) = 3. One can easily see that if P(n) is a permutation then P(P(n)) is a permutation as well. In our example (believe us)    It is natural to denote this permutation by P2(n) = P(P(n)). In a general form the defenition is as follows: P(n) = P1(n), Pk(n) = P(Pk-1(n)). Among the permutations there is a very important one — that moves nothing:    It is clear that for every k the following relation is satisfied: (EN)k = EN. The following less trivial statement is correct (we won't prove it here, you may prove it yourself incidentally): Let P(n) be some permutation of an N elements set. Then there exists a natural number k, that Pk = EN. The least natural k such that Pk = EN is called an order of the permutation P.  The problem that your program should solve is formulated now in a very simple manner: "Given a permutation find its order." Input In the first line of the standard input an only natural number N (1 <= N <= 1000) is contained, that is a number of elements in the set that is rearranged by this permutation. In the second line there are N natural numbers of the range from 1 up to N, separated by a space, that define a permutation — the numbers P(1), P(2),…, P(N). Output You should write an only natural number to the standard output, that is an order of the permutation. You may consider that an answer shouldn't exceed 10 9. Sample Input 5 4 1 5 2 3 Sample Output 6 Source Ural State University Internal Contest October'2000 Junior Session

[Submit]   [Go Back]   [Status]   [Discuss]

题目大意：给出1-n的一个排列，a1,a2,...,an,P(1)=a1,P(2)=a2,...,P(n)=an,P(P(1))=P(a1),P(P(2))=P(a2),
...,P(P(n))=P(an).问经过多少次后使得P(1)=1,...,P(n)=n.求最少多少次，是序列变成一个有序的数列。

题解：置换。

其实相当于给出了一种置换的方式。那么我们可以求出每个轮换中数的个数，即一个轮换中交换多少次可以变成最初的样子，然后求出每个轮换的最小公倍数，即为答案。

#include
#include
#include
#define N 10003
using namespace std;
int ans,cnt[N],a[N],maxn,n,use[N];
int gcd(int x,int y)
{
	int r;
	while (y){
		r=x%y;
		x=y; 
		y=r;
	}
	return x;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	 scanf("%d",&a[i]),cnt[a[i]]++,maxn=max(maxn,a[i]);
	for (int i=1;i<=n;i++)  cnt[i]=cnt[i]+cnt[i-1];
	int ans=1;
	for (int i=1;i<=n;i++)
	 if (!use[i]){
	 	int len=0; int j=i;
	 	while (!use[j]){
	 		len++; 
	 		use[j]=1; j=cnt[a[j]];
	 	}
	 	ans=(ans*len)/gcd(ans,len);
	 }
	printf("%d\n",ans);
}