牛客多校第六场 J Heritage of skywalkert 随即互质概率 nth_element（求最大多少项模板）...-优快云博客

本文介绍了一道ACM竞赛题目，要求求解给定数组中所有不同元素对的最大最小公倍数。通过随机生成数据并筛选出最大值进行暴力求解的方法，给出了一段C++实现代码。

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.

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

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;
}

转载于:https://www.cnblogs.com/l609929321/p/9426160.html