HDU - 5392 Infoplane in Tina Town

题目大意：给出一个序列，求变换几次可以回到原来的位置。比如 1 3 2 ，3 不在原来的位置，变到3位置，次数加1，2变到2，次数+1.得到2.。

解题思路：做法就是分解循环长度。然后求下最小公倍数。但是不能直接用lcm求最小公倍数。。我们可以考虑用质数分解来求，即公共的质因子乘每个数本身的质因子。

#include <cctype>
#include <cstdio>
#include <algorithm>
using namespace std;

const long long MOD = 3221225473;
const int MAXN = 3000010;
int A[MAXN], rec[MAXN]; 

namespace IO {
    const static int maxn = 200 << 20;
    static char buf[maxn], *pbuf = buf, *End;
    void init() {
        int c = fread(buf, 1, maxn, stdin);
        End = buf + c;
    }
    int &readint() {
        static int ans;
        ans = 0;
        while (pbuf != End && !isdigit(*pbuf)) pbuf ++;
        while (pbuf != End && isdigit(*pbuf)) {
            ans = ans * 10 + *pbuf - ‘0‘;
            pbuf ++;
        }
        return ans;
    }
}

int main() {
    IO::init();
    int T;
    T = IO::readint();
    while (T--) {
        int n;
        n = IO::readint();
        for (int i = 1; i <= n; i++)
            A[i] = IO::readint();

        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            int pos = i, tmp = 0, next;
            while (A[pos]) {
                next = A[pos];
                A[pos] = 0;
                pos = next;
                tmp++;
            }
            if (tmp != 0)
                rec[cnt++] = tmp;
        }

        long long ans = 1;
        for (int i = 0; i < cnt; i++) {
            int tmp = rec[i];
            for (int j = 2; j * j <= tmp; j++) {
                int num = 0;
                while (tmp % j == 0) {
                    tmp /= j;
                    num++;
                }
                A[j] = max(A[j], num);
            }
            if (tmp > 1) A[tmp] = max(1, A[tmp]);
        }

        for (int i = 2; i <= n; i++)
            while (A[i]--) ans = (ans * i) % MOD;

        printf("%lld\n", ans);
    }
    return 0;
}