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）...-白红宇

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）...

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发布时间：2019-06-29

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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

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

$\text{lcm}$ 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 ≤ 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

22 1 2 35 3 4 8

输出

Case #1: 68516050958Case #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 #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    #include 
                    
                     #include 
                     
                      #define ls (r<<1)#define rs (r<<1|1)#define debug(a) cout << #a << " " << a << endlusing 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;}