数据的随机性

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

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

题意 ： 给你一个一种随机数据的方式，然后按照他的要求去产生一系列的数，问产生的任意两个数的终 lcm 最大是多少？
思路分析 ：注意此题有个特点，就是A, B, C是随机产生的，这就代表了数据是有一定随机性质的，那么通常我们随机两个数是互质的概率是 6/(pi*pi) ，因此只要取个前 100 个最大的数，然后暴力去查一下即可。
　　注意 ： 就是当循环很大的时候，不要用 STL 中的函数，会超时！！！
代码示例：

#define ll unsigned long long

ll n;
unsigned x, y, z;
priority_queue<unsigned int, vector<unsigned int>, greater<unsigned int> >que;
unsigned arr[105];

ll gcd(ll a, ll b){
    return b==0?a:gcd(b, a%b);
}
ll lcm(ll a, ll b){
    return a/gcd(a, b)*b;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    ll kas = 1;
    ll f1, f2;
    
    cin >> t;
    while(t--){
        scanf("%llu%u%u%u", &n, &x, &y, &z);
        while(!que.empty()) que.pop();
        
        for(ll i = 1; i <= n; i++){
            x ^= x<<16;
            x ^= x>>5;
            x ^= x<<1;
            unsigned f = x; x = y; y = z;
            z = f^x^y;
            
            if (que.size() < 100) que.push(z); 
            else {
                unsigned num = que.top();
                if (z > num) {
                    que.pop();
                    que.push(z);
                }
            }
        }
        ll k = 1;
        while(!que.empty()) {
            arr[k++] = que.top(); que.pop();
        }
        ll ans = 0;
        for(ll i = 1; i < k; i++){
            for(int j = i+1; j < k; j++){
                ll x = lcm((ll)arr[i], (ll)arr[j]);
                if (ans < x) ans = x; 
            }
        }
        printf("Case #%llu: %llu\n", kas++,ans);
    }
    return 0;
}