URAL 1024 Permutations（LCM）

题意：已知，可得出 P(1) = 4, P(2) = 1, P(3) = 5，由此可得出 P(P(1)) = P(4) = 2. And P(P(3)) = P(5) = 3，因此。经过k次如上变换，最终可得，输入保证一定有解，求k。

分析：

1、能用数组表示映射就别用map，很耗时

2、序列中的每个数字经过这种变换都一定会变会自身，例如，一开始P(1) = 4，接下来P(P(1)) = P(4) = 2,再接下来P(P(P(1))) = P(P(4)) = P(2) = 1,只需三次，就可以变回自身（P(P(P(1))) =1），对序列中每个数字求次数，最终求这些次数的最小公倍数即可

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) a < b ? a : b

#define Max(a, b) a < b ? b : a

typedef long long ll;

typedef unsigned long long llu;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {, , -, };

const int dc[] = {-, , , };

const double pi = acos(-1.0);

const double eps = 1e-;

const int MAXN =  + ;

const int MAXT =  + ;

using namespace std;

int x[MAXN];

int w[MAXN];

set<int> y;

int gcd(int a, int b){

    return !b ? a : gcd(b, a % b);

}

int N;

int main()

{

    scanf("%d", &N);

    for(int i = ; i <= N; ++i)

    {

        scanf("%d", &x[i]);

        w[i] = x[i];

    }

    for(int i = ; i <= N; ++i)

    {

        int cnt = ;

        while(x[i] != i){

            x[i] = w[x[i]];

            ++cnt;

        }

        y.insert(cnt);

    }

    set<int>::iterator it = y.begin();

    int ans = *it;

    ++it;

    for(; it != y.end(); ++it){

        ans = ans / gcd(ans, *it) * (*it);//求最小公倍数，调用函数耗时

    }

    printf("%d\n", ans);

    return ;

}