链接：https://www.nowcoder.com/acm/contest/144/J
来源：牛客网
skywalkert, the new legend of Beihang University ACM-ICPC Team, retired this year leaving a group of newbies again.
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.
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
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 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 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
输出
复制Case #1: 68516050958 Case #2: 5751374352923604426
分析：由于数据看上去像是随机⽣成的，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。
比赛的时候想到应该是枚举前面最大的多少项，但是因为不知道怎么快速计算前面多少项没有试一发了！！
结束后补题知道了nth_element求数组中最大的前多少项
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef unsigned long long ll;
const ll maxn = 1e7+10;
const ll mod = 998244353;
unsigned x, y, z, a[maxn];
ll n;
unsigned gcd( unsigned a, unsigned b ) {
if( a == 0 ) {
return b;
} else if( b == 0 ) {
return a;
} else {
return gcd( b, a%b );
}
}
unsigned rnd() {
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() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
ll T, t = 1;
scanf("%llu",&T);
while( T -- ) {
scanf("%llu%u%u%u",&n,&x,&y,&z);
for( ll i = 0; i < n; i ++ ) {
a[i] = rnd();
}
ll m = min( (ll)n, (ll)100 );
nth_element( a, a+n-m, a+n );
ll ans = 0;
for( ll i = n-m; i < n; i ++ ) {
for( ll j = i+1; j < n; j ++ ) {
ans = max( ans, (ll)a[i]*a[j]/gcd(a[i],a[j]) );
}
}
printf("Case #%llu: %llu\n", t++, ans);
}
return 0;
}