保存最大的前20项暴力--Heritage of skywalkert

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

Heritage of skywalkert

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 $\text{lcm}(a_i, a_j)$ . $\text{lcm}$ 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

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，只需取出前20（或100）大暴力即可。随机的两个正整数互质的概率为6/pi^2(百分之60左右）。

直接这样写会TLE

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
unsigned a[10000005];
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;
}
ll gcd(ll a,ll b)
{
    while(b)
    {
        ll c=a%b;
        a=b;
        b=c;
    }
    return a;
}
ll get_lcm(ll x,ll y)
{
    ll d = gcd(x, y);
    return x / d * y;
}
int main()
{
    int t,i,j;
    ll ans;
    scanf("%d", &t);
    for (int cas = 1; cas <= t; cas++)
    {
        int n;
        cin >> n >> x >> y >> z;
        for(i=1;i<=n;i++)
        {
            a[i]=tang();
        }
        sort(a+1,a+1+n);
        ans=0;
        for(i=1;i<=20&&i<=n;i++)
        {
            for(j=i+1;j<=20&&j<=n;j++)
            {
                ans=max(ans,get_lcm(a[i],a[j]));
            }
        }
         cout << "Case #" << cas << ": " << ans << endl;
    }
}

如何优化？

由于我们只用到了最大的20项，就从这20项下手。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ull unsigned long long
ull gcd(ull a,ull b){
    return b?gcd(b,a%b):a;
}
ull get_lcm(ull a,ull b)
{
    return a/gcd(a,b)*b;
}
unsigned a[30];
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;
}
/*ll gcd(ll a,ll b)
{
    while(b)
    {
        ll c=a%b;
        a=b;
        b=c;
    }
    return a;
}
ll get_lcm(ll x,ll y)
{
    ll d = gcd(x, y);
    return x / d * y;
}*/
int main()
{
    int t,i,j,k,pos,temp;
    ull ans;
    scanf("%d", &t);
    for (int cas = 1; cas <= t; cas++)
    {
        int n;
        cin >> n >> x >> y >> z;
        pos=1;
        for(i=1;i<=n;i++)
        {
            if(pos<=20)
            {
                a[pos]=tang();
                pos++;
                sort(a+1,a+pos);
            }
            else
            {
               temp=tang();
               j=1;
               while(a[j]<temp&&j<pos) j++;
               for(k=1;k<j-1;k++)
               {
                   a[k]=a[k+1];
               }
               if(j!=1)
               a[j-1]=temp;
            }
        }
        ans=0;
        for(i=1;i<pos&&i<=n;i++)
        {
            for(j=i+1;j<pos&&j<=n;j++)
            {
                ans=max(ans,get_lcm(a[i],a[j]));
            }
        }
         cout << "Case #" << cas << ": " << ans << endl;
    }
}

这个题姿势不好还真不好过，找错找了一上午，最后把注释掉的换掉就过了。