zoukankan      html  css  js  c++  java
  • HDU6025 Coprime Sequence —— 前缀和 & 后缀和

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6025


    Coprime Sequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 666    Accepted Submission(s): 336


    Problem Description
    Do you know what is called ``Coprime Sequence''? That is a sequence consists of n positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1.
    ``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.
     

    Input
    The first line of the input contains an integer T(1T10), denoting the number of test cases.
    In each test case, there is an integer n(3n100000) in the first line, denoting the number of integers in the sequence.
    Then the following line consists of n integers a1,a2,...,an(1ai109), denoting the elements in the sequence.
     

    Output
    For each test case, print a single line containing a single integer, denoting the maximum GCD.
     

    Sample Input
    3 3 1 1 1 5 2 2 2 3 2 4 1 2 4 8
     

    Sample Output
    1 2 2
     



    题解:

    l[i]为前i个数的gcd, r[i]为后i个数的gcd。

    假设被删除的数的下标为i, 则 删除该数后的gcd为: gcd(l[i-1], r[i+1]), 枚举i,取最大值。



    学习之处:

    当提到在序列里删除一段连续的数时,可以用前缀和+后缀和

    例如:http://blog.csdn.net/dolfamingo/article/details/71001021



    代码如下:

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    const double eps = 1e-6;
    const int INF = 2e9;
    const LL LNF = 9e18;
    const int mod = 1e9+7;
    const int maxn = 1e5+10;
    
    int n;
    int a[maxn], l[maxn], r[maxn];
    
    int gcd(int a, int b)
    {
        return b==0?a:(gcd(b,a%b));
    }
    
    void solve()
    {
        scanf("%d",&n);
        for(int i = 1; i<=n; i++)
            scanf("%d",&a[i]);
    
        l[1] = a[1]; r[n] = a[n];
        for(int i = 2; i<=n; i++)
            l[i] = gcd(l[i-1], a[i]);
        for(int i = n-1; i>=1; i--)
            r[i] = gcd(r[i+1], a[i]);
    
        int ans = 1;
        l[0] = a[2]; r[n+1] = a[n-1];
        for(int i = 1; i<=n; i++)
            ans = max(ans, gcd(l[i-1], r[i+1]) );
        cout<<ans<<endl;
    }
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            solve();
        }
        return 0;
    }
    


  • 相关阅读:
    收藏CSS经典技巧
    理解这26句话将不再烦恼
    包转发率得计算和背板带宽的计算
    mysql 建表 AUTO_INCREMENT , 数据类型 VARCHAR
    Linux Wine with *.bat *.exe ( Photoshop and etc.. )
    [转载]expect spawn、linux expect 用法小记
    sqlmap.py Database injection and hak
    xls===>csv tables===via python ===> sqlite3.db
    sftp 服务器外网访问设置
    vsftp FTP服务器外网访问设置
  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/7538680.html
Copyright © 2011-2022 走看看