poj 2369 Permutations - 数论
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
Output
Sample Input
5
4 1 5 2 3
Sample Output
6
题目大意是讲给出一个置换,定义它和它自己的合成运算,问它和它自己进行多少次合成运算后又变回了自己。
根据置换的知识,任何一个置换都可以表示成轮换
然后根据人生的经验和数学的直觉,循环周期等于当置换表示成轮换的合成的形式时,每个轮换中元素的个数的最小公倍数(每次每个轮换往前转一次,如果还不能理解,出门左转<组合数学>)。
Code
/**
* poj
* Problem#2369
* Accepted
* Time:16ms
* Memory:692k
*/
#include<iostream>
#include<fstream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<ctime>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#ifndef WIN32
#define AUTO "%lld"
#else
#define AUTO "%I64d"
#endif
using namespace std;
typedef bool boolean;
#define inf 0xfffffff
#define smin(a, b) (a) = min((a), (b))
#define smax(a, b) (a) = max((a), (b))
template<typename T>
inline boolean readInteger(T& u) {
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-') {
aFlag = -;
x = getchar();
}
for(u = x - ''; isdigit((x = getchar())); u = u * + x - '');
u *= aFlag;
ungetc(x, stdin);
return true;
} template<typename T>
T gcd(T a, T b) {
if(b == ) return a;
return gcd(b, a % b);
} int n;
int *f; inline void init() {
readInteger(n);
f = new int[(const int)(n + )];
for(int i = ; i <= n; i++)
readInteger(f[i]);
} int lcm = ;
boolean *visited;
inline void solve() {
visited = new boolean[(const int)(n + )];
memset(visited, false, sizeof(boolean) * (n + ));
for(int i = ; i <= n; i++) {
if(!visited[i]) {
int c = , j = i;
while(!visited[j]) {
visited[j] = true;
j = f[j], c++;
}
lcm = lcm / gcd(c, lcm) * c;
}
}
printf("%d", lcm);
} int main() {
init();
solve();
return ;
}
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