Coprime Sequence HDU - 6025

最新推荐文章于 2019-11-16 21:13:50 发布

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最新推荐文章于 2019-11-16 21:13:50 发布

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分类专栏： # 动态规划（dp）

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本文介绍了一种算法问题，即给定一个整数序列，在移除一个元素后找到该序列中剩余元素的最大公约数的最大值。文章提供了问题的背景、解决思路及AC代码实现。

Coprime Sequence

HDU - 6025

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nn 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.

InputThe first line of the input contains an integer T(1≤T≤10)T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000)n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of nn integers a1,a2,...,an(1≤ai≤109)a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.OutputFor each test case, print a single line containing a single integer, denoting the maximum GCD.Sample Input

Sample Output

1
2
2

题意：已知长度为n的序列，问删除其中一个数后，这个新序列的最大公约数最大能为多少。

思路：对于第i个数，其前面的gcd和其后面的gcd都是固定的，先按顺序分别从前向后再后向前分别记录每个子串的gcd并记录，最后模拟删除第i个数后对其前后求最大公约数，比较出最大的即可。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn 100010
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 100000000
using namespace std;

int a[maxn],dp1[maxn],dp2[maxn];
int gcd(int x,int y)
{
    if(y==0) return x;
    return gcd(y,x%y);
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
        }
        dp1[1] = a[1];
        dp2[n] = a[n];
        for(int i=2;i<=n;++i){
            dp1[i] = gcd(dp1[i-1],a[i]);
        }
        for(int i=n-1;i>=1;--i){
            dp2[i] = gcd(dp2[i+1],a[i]);
        }
        int MAX = max(dp2[2],dp1[n-1]); //判断去头或者去尾时最大公倍数
        for(int i=2;i<=n-1;++i){
            MAX = max(MAX,gcd(dp1[i-1],dp2[i+1]));
        }
        cout<<MAX<<endl;
    }
    return 0;
}