HDU6025 Coprime Sequence【前缀GCD+后缀GCD】

 

Coprime Sequence

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1415    Accepted Submission(s): 690

 

 

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(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of n integers a1,a2,...,an(1≤ai≤109), 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

 

 

Source

2017中国大学生程序设计竞赛 - 女生专场

 

 

 

 

问题链接:HDU6025 Coprime Sequence。

问题简述

  去除数列中的一个数字,使去除后数列中所有数字的gcd尽可能大。

问题分析

  这是一个计算GCD的问题。

程序说明

  数组prefixgcd[],对于prefixgcd[i]=g,g为a[0]-a[i]的GCD,称为前缀GCD。

  数组suffixgcd[],对于suffixgcd[i]=g,g为a[i]-a[n-1]的GCD,称为后缀GCD。

  有了这两个GCD值的数组,那么去掉a[i]的GCD为gcd(prefixgcd[i - 1], suffixgcd[i + 1]),从中找出最大值即可。

 

题记:(略)

 

 

AC的C++语言程序如下:

 

/* HDU6025 Coprime Sequence */

#include 

using namespace std;

const int N = 100000;
int a[N], prefixgcd[N], suffixgcd[N];

inline int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main()
{
    int t, n;

    cin >> t;
    while(t--) {
        cin >> n;

        for(int i=0; i> a[i];

        // 计算前缀GCD
        prefixgcd[0] = a[0];
        for(int i=1; i=0; i--)
            suffixgcd[i] = gcd(a[i], suffixgcd[i + 1]);

        int ans = max(suffixgcd[1], prefixgcd[n - 2]);
        for(int i=1; i

 

 

 

 

 

 

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