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 ≤ 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：n个数字，给出x，y，z，按照给出的程序返回第i个值为a[i]，问这n个数字任意两个数字的lcm的最大值
思路：题解很玄学，最大的一百个数字，暴力两两求lcm，取最大即可。至于为什么？没有为什么！
注意给出的程序用的是unsigned int，不要随便改！！因为在左移右移的时候会去掉抄出范围的部分，改了类型，数字长度范围改变了，得到的数字是不对的，并且，ans定义不能为long long，要用unsigned long long，因为求lcm的数字是unsigned int型的，比int大，所以long long 会爆掉。
这个代码中用到了函数nth_element（a,a+b,a+n），a是数组的头地址，c+n是数组的尾地址，这个函数会使得数组b位置前小于a[b]这个值，而后面的大于这个值，但是前后数字无序，复杂度为O（n）
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;
typedef unsigned long long ll;
unsigned x,y,z;
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 a[10000007];
int main(){
int t;
scanf("%d",&t);
int c=1;
while(t--){
int n;
cin>>n>>x>>y>>z;
for(int i=1;i<=n;i++){
a[i]=tang();
}
int len=max(1,n-100);
nth_element(a+1,a+1+len,a+1+n);
ll ans=0;
for(int i=len;i<=n;i++){
for(int j=len;j<=n;j++){
if(i==j)continue;
ans=max(ans,(ll)a[i]/__gcd(a[i],a[j])*a[j]);
}
}
cout<<"Case #"<<c++<<": "<<ans<<endl;
}
return 0;
}