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  • 牛客网暑期ACM多校训练营(第六场)J 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  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

    题意: 读题只读最后一句话系列。t组样例,t不超过50。每组样例四个数字n,a,b,c,n代表序列有n个数字,不超过5组样例n大于2e6,n<1e7。a,b,c在推出序列的过程中使用,数据范围unsigned int
    按照题意中给出的代码,每运行一次得到的就是a[i],最后要求选出两个数字使其LCM最大,输出最大的LCM。
    做法:丧心病狂,不是第一次遇见这样的解法了,也就是保留前100大,然后暴力求LCM即可。
    注意******* 题目上说了a,b,c都是unsigned int类型的,不能开成unsigned long long类型的,不正常取舍答案会错误。
    nth_element(a,a+k,a+n,cmp);可以得到前k小或者前k大,自定义排序方式
    代码如下:
    #include<stdio.h>
    #include<iostream>
    #include<algorithm>
    
    using namespace std;
    
    const int maxn = 10000007;
    
    int t;
    int n;
    unsigned int x , y , z;
    unsigned long long num[maxn];
    
    unsigned long long gcd(unsigned long long a , unsigned long long b)
    {
        unsigned long long c;
        c = a%b;
        while( c )
        {
            a = b;
            b = c;
            c = a%b;
        }
        return b;
    }
    
    unsigned int solve()
    {
        unsigned int tt;
        x ^= x<<16;
        x ^= x>>5;
        x ^= x<<1;
        tt = x;
        x = y;
        y = z;
        z = tt^x^y;
        return z;
    }
    
    bool cmp(unsigned long long a , unsigned long long b)
    {
        return a>b;
    }
    
    int main()
    {
        scanf("%d" , &t);
        for(int cas=1; cas<=t; cas++)
        {
            scanf("%d%u%u%u" , &n , &x , &y , &z);
            for(int i=0; i<n; i++)
            {
                num[i] = solve();
            }
            int k = min(100 , n);   ///取前100大
            nth_element(num , num+k , num+n , cmp);
            unsigned long long ans;
            ans = 0;
            for(int i=0; i<k; i++)
            {
    //            printf("%llu
    " , num[i]);
                for(int j=i+1; j<k; j++)
                {
                    ans = max(ans , num[i]*num[j]/gcd(num[i],num[j]));
    //                 printf("%d..%d..%llu..%llu..%llu..%llu..
    " , i , j , num[i] , num[j] , num[i]*num[j]/gcd(num[i],num[j]) , ans);
                }
            }
            printf("Case #%d: %llu
    " , cas , ans);
    
        }
    
    
        return 0;
    }
    
    /*
    337929
    608269
    1351708
    64488027082
    85984357633
    
    405514
    675854
    1351708
    1384776332
    2769433858
    
    0..1..405514..675854..137034129478..137034129478..
    0..2..405514..1351708..274068258956..274068258956..
    0..3..405514..1384776332..280773094747324..280773094747324..
    0..4..405514..2769433858..561522100746506..561522100746506..
    1..2..675854..1351708..1351708..561522100746506..
    1..3..675854..1384776332..467953311543764..561522100746506..
    1..4..675854..2769433858..935866475332366..935866475332366..
    2..3..1351708..1384776332..467953311543764..935866475332366..
    2..4..1351708..2769433858..1871732950664732..1871732950664732..
    3..4..1384776332..2769433858..1917523229798924428..1917523229798924428..
    */




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  • 原文地址:https://www.cnblogs.com/Flower-Z/p/9643476.html
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