HDU5392 Infoplane in Tina TownLCM

Posted 2021-09-04 海岛Blog

Infoplane in Tina Town
Time Limit: 14000/7000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2713 Accepted Submission(s): 730

Problem Description
There is a big stone with smooth surface in Tina Town. When people go towards it, the stone surface will be lighted and show its usage. This stone was a legacy and also the center of Tina Town’s calculation and control system. also, it can display events in Tina Town and contents that pedestrians are interested in, and it can be used as public computer. It makes people’s life more convenient (especially for who forget to take a device).

Tina and Town were playing a game on this stone. First, a permutation of numbers from 1 to n were displayed on the stone. Town exchanged some numbers randomly and Town recorded this process by macros. Town asked Tine,”Do you know how many times it need to turn these numbers into the original permutation by executing this macro? Tina didn’t know the answer so she asked you to find out the answer for her.

Since the answer may be very large, you only need to output the answer modulo 3∗2³⁰+1=3221225473 (a prime).

Input
The first line is an integer T representing the number of test cases. T≤5

For each test case, the first line is an integer n representing the length of permutation. n≤3∗10⁶

The second line contains n integers representing a permutation A1…An. It is guaranteed that numbers are different each other and all Ai satisfies ( 1≤Ai≤n ).

Output
For each test case, print a number ans representing the answer.

Sample Input
2
3
1 3 2
6
2 3 4 5 6 1

Sample Output
2
6

Source
BestCoder Round #51 (div.2)

问题链接：HDU5392 Infoplane in Tina Town
问题简述：（略）
问题分析：LCM问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* HDU5392 Infoplane in Tina Town */

#include <bits/stdc++.h>

using namespace std;

const long long MOD = 3221225473;
const int N = 3e6 + 1;
int a[N], vis[N], num[N];

int main()
{
    int t, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

        memset(vis, 0, sizeof(int) * (n + 1));
        memset(num, 0, sizeof(int) * (n + 1));
        for (int i = 1; i <= n; i++) {
            if (vis[i] == 0) {
                // 计算循环长度
                int cnt = 0, k = i;
                while (vis[k] == 0)
                    vis[k] = 1, cnt++, k = a[k];
                for (int j = 2; j * j <= cnt; j++) {
                    int sum = 0;
                    if (cnt % j == 0)
                        while (cnt % j == 0) cnt /= j, sum++;
                    num[j] = max(num[j], sum);
                }
                if (cnt > 1) num[cnt] = max(num[cnt], 1);
            }
        }

        long long ans = 1;
        for (int i = 2; i <= n; i++)
            while (num[i]--) ans = ans * i % MOD;

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

    return 0;
}