（第六场）Heritage of skywalkert 【玄学】

题目链接：https://www.nowcoder.com/acm/contest/144/J

标题：J、Heritage of skywalkert

| 时间限制：1 秒 | 内存限制：256M

skywalkert, the new legend of Beihang University ACM-ICPC Team, retired this year leaving a group of newbies again. Rumor has it that he left a heritage when he left, and only the one who has at least 0.1% IQ(Intelligence Quotient) of his can obtain it. To prove you have at least 0.1% IQ of skywalkert, you have to solve the following problem: Given n positive integers, for all (i, j) where 1 ≤ i, j ≤ n and i ≠ j, output the maximum value

among means the Lowest Common Multiple.

输入描述:

The input starts with one line containing exactly one integer t which is the number of test cases. (1 ≤ t ≤ 50)

For each test case, the first line contains four integers n, A, B, C. (2 ≤ n ≤ 107, A, B, C are randomly selected in unsigned 32 bits integer range)

The n integers are obtained by calling the following function n times, the i-th result of which is ai, and we ensure all ai > 0. Please notice that for each test case x, y and z should be reset before being called.

No more than 5 cases have n greater than 2 x 106.

输出描述:

For each test case, output "Case #x: y" in one line (without quotes), where x is the test case number (starting from 1) and y is the maximum lcm.

示例 1

输入

2 1 2 3

5 3 4 8

输出

Case #1: 68516050958

Case #2: 5751374352923604426

题意概括：

按照给出的函数打出 N 个数，求其中两个数的最小公倍数的最大值。

官方题解：

由于数据看上去像是随机⽣生成的，只需要选出前 100 大的数平方暴力即可。随机两个正整数互质的概率为 6/(π^2) = 0.608...

解题思路：

nth_element（）选出前100大的数，暴力前 100 大的数取最大的最小公倍数。

AC code：

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cmath>

 #define INF 0x3f3f3f3f

 #define ll unsigned long long

 using namespace std;

 const int MAXN = 1e7+;

 ll num[MAXN];

 unsigned int N, A, B, C;

 unsigned int x, y, z;

 ll gcd(ll x, ll y)

 {

     return y==?x:gcd(y, x%y);

 }

 bool cmp(ll x, ll y)

 {

     return x>y;

 }

 unsigned tang()

 {

     unsigned t;

     x ^= x<<;

     x ^= x>>;

     x ^= x<<;

     t = x;

     x = y;

     y = z;

     z = t^x^y;

     return z;

 }

 int main()

 {

     int T_case;

     scanf("%d", &T_case);

     int cnt = ;

     while(T_case--)

     {

         scanf("%u%u%u%u", &N, &A, &B, &C);

         x = A, y = B, z = C;

         for(int i = ; i < N; i++)

         {

             num[i] = tang();

         }

         unsigned int k = min(100u, N);

         nth_element(num, num+k, num+N, cmp);

         ll res = ;

         for(int i = ; i < k; i++)

             for(int j = i+; j < k; j++)

         {

             res = max(res, num[i]*num[j]/gcd(num[i],num[j]));

         }

         printf("Case #%d: %llu\n", ++cnt, res);

     }

     return ;

 }