2018牛客多校训练----Heritage of skywalkert

链接：https://www.nowcoder.com/acm/contest/144/J
来源：牛客网
 

题目描述

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.

输入

2
2 1 2 3
5 3 4 8

输出

Case #1: 68516050958
Case #2: 5751374352923604426

 输入n,A,B,C,a[i]是上面的程序跑i次得出的，求a数组中任意两个数最小公倍数的最大值。

思路：

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ULL;
const int N=1e7+5;

int T,n;
unsigned x,y,z,a[N];

unsigned read()
{
    unsigned t;
    x^=x<<16;
    x^=x>>5;
    x^=x<<1;
    t=x;
    x=y;
    y=z;
    z=t^x^y;
    return z;
}

int main()
{
    scanf("%d",&T);
    int Case=0;
    while (T--)
    {
        scanf("%d%u%u%u",&n,&x,&y,&z);
        for (int i=1;i<=n;i++)
            a[i]=read();
        int m=max(0,n-100);
        nth_element(a+1,a+m+1,a+n+1);
        ULL ans=0;
        for (int i=m+1;i<=n;i++)
            for (int j=i+1;j<=n;j++)
                ans=max(ans,(ULL)a[i]/__gcd(a[i],a[j])*a[j]);
        printf("Case #%d: %llu\n",++Case,ans);
    }
    return 0;
}