牛客网暑期ACM多校训练营（第六场）J-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 $\text{lcm}(a_i, a_j)$ . $\text{lcm}$ 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

题意:给出初始的A，B，C调用tang(）函数n次求n个z，求这n个数中的两个数lcm的最大值。

思路：比赛的时候一直以为是有什么奇怪的结论可以化简，打算打表找规律，后来最近被随机数题坑傻了，但是也不知道怎么写，官方给的思路就是保留最大的100项，然后暴力求就能过。学到了新的，很好用但是之前没怎么用过的函数nth_element(start, start+n, end)百度百科链接，总体来说，这个题，毒性不小...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>

using namespace std;

#define inf 0x3f3f3f3f

typedef unsigned long long ULL;

const int maxn=1e9+7;

unsigned x,y,z;
unsigned data[maxn];

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

unsigned gcd(unsigned a, unsigned b)
{
    if(b == 0) return a;
    return gcd(b,a%b);
}

int main()
{
    int t;
    int n;
    scanf("%d",&t);
    for (int cases = 1; cases <= t; ++cases)
    {
        scanf("%d",&n);
        scanf("%u %u %u",&x,&y,&z);
        for (int i = 1; i <= n; ++i) data[i]=tang();
        int m=max(0,n-100);
        nth_element(data+1,data+m+1,data+n+1);
        unsigned long long ans=0;
        for (int i = m; i <= n ; ++i)
        {
            for (int j = i+1; j <= n ; ++j)
            {
                ans=max(ans,1ULL*data[i]/gcd(data[i],data[j])*data[j]);
            }
        }
        printf("Case #%d: %llu\n",cases,ans);

    }
    return 0;
}