Codeforces 742 C Arpa's loud Owf and Mehrdad's evil plan

题目地址：http://codeforces.com/contest/742/problem/C
题意：我看了好多别人的博客的对于题意的理解，结合他们的代码我认为这道题的意思应该是每个数要都能经过t步以后到达一个点，再经过t步到达起点。
思路：看这些数能否构成一个或多个环（包括自环），如果存在不能成环的数就输出-1，否则一定存在一个最小的t。

如果当前环中元素的个数为奇数，则 x->x->x 当前的t=元素个数
如果当前环中元素的个数为偶数，则 x->y->x （y为中间的数）当前的t=元素个数/2
然后再把每个环的 t 求出他们的最小公倍数就是最小的 t，对于每个环都满足的t。

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define LL long long
#define N 110
#define M 50010
#define inf 0x3f3f3f3f
using namespace std;
const LL mod = 1e9 + 7;
const double eps = 1e-9;
LL num[N], pre[N];
LL gcd(LL a, LL b) {return b == 0 ? a : gcd(b, a%b);
}
LL lcm(LL a, LL b) {return a*b / gcd(a, b);
}
int main() {cin.sync_with_stdio(false);LL n, flag, ans, cnt, sum;while (cin >> n) {memset(pre, 0, sizeof(pre));for (int i = 1; i <= n; i++) {cin >> num[i];}flag = 1;sum = 0;for (int i = 1; i <= n&&flag; i++) {if (!pre[i]) {ans = i;cnt = 0;while (pre[num[ans]] == 0) {pre[num[ans]] = ans;ans = num[ans];cnt++;}if (pre[ans] == 0 || ans != i) {flag = 0;}else {if (cnt % 2) {if (sum == 0) {sum = cnt;}else {sum = lcm(sum, cnt);}}else {if (sum == 0) {sum = cnt / 2;}else {sum = lcm(sum, cnt / 2);}}}}}if (flag) {cout << sum << endl;}else {cout << -1 << endl;}}return 0;
}

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