（第六场）Heritage of skywalkert 【玄学】

原创于 2018-08-11 12:30:00 发布 · 191 阅读

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本篇博客解析了Heritageofskywalkert题目，介绍了如何通过选取前100个最大数值来求解任意两数间最大最小公倍数的问题。利用nth_element算法优化并实现快速求解。

题目链接：https://www.nowcoder.com/acm/contest/144/J

标题：J、Heritage of skywalkert

| 时间限制：1 秒 | 内存限制：256M

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.

示例 1

输入

2 1 2 3

5 3 4 8

输出

Case #1: 68516050958

Case #2: 5751374352923604426

题意概括：

按照给出的函数打出 N 个数，求其中两个数的最小公倍数的最大值。

官方题解：

由于数据看上去像是随机⽣生成的，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^2) = 0.608...

解题思路：

nth_element（）选出前100大的数，暴力前 100 大的数取最大的最小公倍数。

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cmath>
 5 #define INF 0x3f3f3f3f
 6 #define ll unsigned long long
 7 using namespace std;
 8 
 9 const int MAXN = 1e7+10;
10 
11 ll num[MAXN];
12 unsigned int N, A, B, C;
13 unsigned int x, y, z;
14 
15 ll gcd(ll x, ll y)
16 {
17     return y==0?x:gcd(y, x%y);
18 }
19 
20 bool cmp(ll x, ll y)
21 {
22     return x>y;
23 }
24 
25 unsigned tang()
26 {
27     unsigned t;
28     x ^= x<<16;
29     x ^= x>>5;
30     x ^= x<<1;
31     t = x;
32     x = y;
33     y = z;
34     z = t^x^y;
35     return z;
36 }
37 
38 int main()
39 {
40     int T_case;
41     scanf("%d", &T_case);
42     int cnt = 0;
43     while(T_case--)
44     {
45 
46         scanf("%u%u%u%u", &N, &A, &B, &C);
47         x = A, y = B, z = C;
48         for(int i = 0; i < N; i++)
49         {
50             num[i] = tang();
51         }
52         unsigned int k = min(100u, N);
53         nth_element(num, num+k, num+N, cmp);
54         ll res = 0;
55         for(int i = 0; i < k; i++)
56             for(int j = i+1; j < k; j++)
57         {
58             res = max(res, num[i]*num[j]/gcd(num[i],num[j]));
59         }
60         printf("Case #%d: %llu\n", ++cnt, res);
61     }
62     return 0;
63 }