# HDOJ 5392-Infoplane in Tina Town

HDOJ 5392-Infoplane in Tina Town

Infoplane in Tina Town
Time Limit: 14000/7000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1636 Accepted Submission(s): 385

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)

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60=2×2×3×5

18=2×3×3

lcm(60,18)=180=2×2×3×3×5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef unsigned long long ULL;
const int kMaxN = 3000005, kNumPrime = 269;
const ULL kMod = 3221225473;
bool visited[kMaxN];
int permutation[kMaxN];
int prime[kNumPrime] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723 };
int prime_times[kMaxN];

//求x的所有质因子及其频率
void GetPrimeFactor(int x)
{
for (int i = 0; i < kNumPrime; i++)
{
int tmp = 0;
while (x%prime[i] == 0)
{
tmp++;
x /= prime[i];
}
if (tmp > prime_times[prime[i]])
prime_times[prime[i]] = tmp;
if (x == 1)
return;
}
prime_times[x]++;
}
//快速幂取模x^ymod(kMod)
ULL QuickPow(ULL x, ULL y)
{
ULL ans = 1;
while (y)
{
if (y & 1)
ans = (ans*x)%kMod;
x = (x*x) % kMod;
y >>= 1;
}
return ans;
}
int main()
{
int t, n, len;
ULL ans;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
memset(visited, false, kMaxN);
memset(prime_times, 0, kMaxN);
for (int i = 1; i <= n; i++)
scanf("%d", &permutation[i]);
for (int i = 1; i <= n; i++)
{
if (!visited[i])
{
visited[i] = true;
int j = permutation[i];
len = 1;
while (j != i)
{
visited[j] = true;
j = permutation[j];
len++;
}
GetPrimeFactor(len);
}
}
ans = 1;
for (int i = 2; i < n; i++)
if (prime_times[i] != 0)
ans = (ans*QuickPow(i, prime_times[i])) % kMod;
printf("%I64un", ans);
}
return 0;
}