J Heritage of skywalkert

 牛客网暑期ACM多校训练营（第六场）  J  Heritage of skywalkert

题目：

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

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

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 ≤ 10

⁷

, 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 a_i, and we ensure all a_i > 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 10

⁶

.

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

复制

2
2 1 2 3
5 3 4 8

输出

复制

Case #1: 68516050958
Case #2: 5751374352923604426

思路：　

　 调用题目的给的函数tang(), 生成数a[1]到a[n], 然后调用STL的库函数nth_element(原理是快排，也可以自己手搓) ，选出前100大的数，然后暴力选两个数，使得最小公倍数最大。

附: 

  关于STL中的nth_element()方法的使用：通过调用nth_element(start, start+n, end) 方法可以使第n大元素处于第n位置（从0开始,其位置是下标为 n的元素），并且比这个元素小的元素都排在这个元素之前，比这个元素大的元素都排在这个元素之后，但不能保证他们是有序的。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn =1e7+20;
unsigned x,y,z,A,B,C;
typedef unsigned long long ull;

unsigned a[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;
}

int main()
{
    int t,n;
    cin>>t;
    for(int cas=1;cas<=t;cas++){
        cin>>n>>A>>B>>C;
        x=A,y=B,z=C;
        for(int i=1;i<=n;i++)
            a[i]=tang();

        vector<unsigned> vec;
        int num=min(n,100);
        nth_element(a+1,a+1+n-num,a+1+n);  ///只要排一次

        /// 选出前一百大的数
        for(int k=1;k<=num;k++){
            vec.push_back(a[n+1-k]);
        }

       ///暴力找lcm最大
        ull  ans=0;
        for(int i=0;i<vec.size();i++){
            for(int j=i+1;j<vec.size();j++){
                ull gcd = __gcd((ull)vec[i],(ull)vec[j]);
                ull tp=vec[i]/gcd*vec[j];
                ans = max(ans,tp);
            }
        }
        printf("Case #%d: ",cas);
        cout<<ans<<endl;
    }


    return 0;
}