【CODECHEF】【phollard rho + miller_rabin】The First Cube

All submissions for this problem are available.

Read problems statements in Mandarin Chinese and Russian.

This problem's statement is really a short one.

You are given an integer S. Consider an infinite sequence S, 2S, 3S, ... . Find the first number in this sequence that can be represented as Q³, where Q is some positive integer number. As the sought number could be very large, please print modulo (10⁹ + 7).

The number S will be given to you as a product of N positive integer numbers A₁, A₂, ..., A_N, namely S = A₁ * A₂ * ... * A_N

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.

The first line of each test case contains an integer N.

Then there is a line, containing N space separated integers, denoting the numbers A₁, A₂, ..., A_N.

Output

For each test case, output a single line containing the first term of the sequence which is the perfect cube, modulo 10⁹+7.

Constraints

• 1 ≤ T ≤ 10
• (Subtask 1): N = 1, 1 ≤ S ≤ 10⁹ - 15 points.
• (Subtask 2): N = 1, 1 ≤ S ≤ 10¹⁸ - 31 point.
• (Subtask 3): 1 ≤ N ≤ 100, 1 ≤ A_i ≤ 10¹⁸ - 54 points.

Example

Input:

2

2

2 2

2

2 3

Output:

8

216

Explanation

Example case 1. First few numbers in the infinite sequence 4, 8, 12, 16, 20, , etc. In this sequence, 8 is the first number which is also a cube (as 2³ = 8).

Example case 2. First few numbers in the infinite sequence 6, 12, 18, 24, , etc. In this sequence, 216 is the first number which is also a cube (as 6³ = 216).

【分析】

挺模板的东西，就当复习一下了。

  1 /*

  2 宋代李冠

  3 《蝶恋花·春暮》

  4 遥夜亭皋闲信步。

  5 才过清明，渐觉伤春暮。

  6 数点雨声风约住。朦胧淡月云来去。

  7 桃杏依稀香暗渡。

  8 谁在秋千，笑里轻轻语。

  9 一寸相思千万绪。人间没个安排处。 

 10 */

 11 #include <cstdio>

 12 #include <cstring>

 13 #include <algorithm>

 14 #include <cmath>

 15 #include <queue>

 16 #include <vector>

 17 #include <iostream>

 18 #include <string>

 19 #include <ctime>

 20 #include <map>

 21 #define LOCAL

 22 const int MAXN = 105 + 10;

 23 const long long MOD = 1000000007;

 24 const double Pi = acos(-1.0);

 25 long long G = 15;//原根 

 26 const int MAXM = 60 * 2 + 10; 

 27 using namespace std;

 28 typedef long long ll;

 29 ll read(){

 30    ll flag = 1, x = 0;

 31    char ch;

 32    ch = getchar();

 33    while (ch < '1' || ch > '9') {if (ch == '-') flag = -1; ch = getchar();}

 34    while (ch >= '0' && ch <= '9') {x = x * 10 + (ch - '0'); ch = getchar();}

 35    return x * flag;

 36 }

 37 map<ll, int>Num;//记录个数

 38 ll data[MAXN]; 

 39 ll n;

 40 

 41 ll mul(ll a, ll b, ll c){//又要用快速乘QAQ

 42    if (b == 0) return 0ll;

 43    if (b == 1) return a % c;

 44    ll tmp = mul(a, b / 2, c);

 45    if (b % 2 == 0) return (tmp + tmp) % c;

 46    else return (((tmp + tmp) % c) + (a % c)) % c;

 47 }

 48 ll pow(ll a, ll b, ll c){

 49    if (b == 0) return 1ll;

 50    if (b == 1) return a % c;

 51    ll tmp = pow(a, b / 2, c);

 52    if (b % 2 == 0) return mul(tmp, tmp, c);

 53    else return mul(mul(tmp, tmp, c), a, c);

 54 }

 55 bool Sec_check(ll a, ll b, ll c){

 56      ll tmp = pow(a, b, c);

 57      if (tmp != 1 && tmp != (c - 1)) return 0;

 58      if (tmp == (c - 1) || (b % 2 != 0)) return 1; 

 59      return Sec_check(a, b / 2, c);

 60 }

 61 //判断n是否是素数 

 62 bool miller_rabin(ll n){

 63      int cnt = 20;

 64      while (cnt--){

 65            ll a = (ll)rand() % (n - 1) + 1;

 66            if (!Sec_check(a, n - 1, n)) return 0;

 67      } 

 68      return 1;

 69 }

 70 ll gcd(ll a, ll b){return b == 0ll ? a : gcd(b, a % b);}

 71 ll pollard_rho(ll a, ll c){

 72    ll i = 1, k = 2;

 73    ll x, y, d;

 74    x = (ll)((double)(rand() / RAND_MAX) * (a - 2) + 0.5) + 1ll; 

 75    y = x;

 76    while (1){

 77          i++;

 78          x = (mul(x, x, a) % a + c) % a;

 79          d = gcd(y - x + a, a);

 80          if (d > 1 && d < a) return d;

 81          if (y == x) return a;//失败

 82          if (i == k){

 83             k <<= 1;

 84             y = x;

 85          }

 86    }

 87 }

 88 void find(ll a, ll c){

 89      if (a == 1) return;

 90      if (miller_rabin(a)){

 91         Num[a]++;

 92         return;

 93      }

 94      ll p = a;

 95      while (p >= a) pollard_rho(a, c--);

 96      pollard_rho(p, c);

 97      pollard_rho(a / p, c);

 98 }

 99 void init(){

100      Num.clear();

101      scanf("%d", &n);

102      for (int i = 1; i <= n; i++) {

103          data[i] = read();

104          find(data[i], 15000);

105      }

106 } 

107 void work(){

108      ll Ans = 1;

109      for (int i = 1; i <= n; i++) Ans = (Ans * data[i]) % MOD;

110      for (map<ll, int>::iterator it = Num.begin(); it != Num.end(); it++){

111          it->second %= 3;

112          if (it->second){

113             for (int i = it->second; i < 3; i++) Ans = (Ans * ((it->first) % MOD)) % MOD;

114          }

115      }

116      printf("%lld\n", Ans);

117 }

118 

119 int main(){

120     int T;

121     

122     scanf("%d", &T);

123     while (T--){

124           init();

125           work();

126     }

127     return 0;

128 }

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