HDU6025 Coprime Sequence (思路题）

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 785    Accepted Submission(s): 411

Problem Description

Do you know what is called ``Coprime Sequence''? That is a sequence consists ofn 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

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

 

Source

2017中国大学生程序设计竞赛 - 女生专场

 

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题意

n个数，最大公约数为一，求去掉一个数，求出最大公约数

思路

从左往右，找出前几个数的最大公约数  gleft[i]

从右往左，找出后几个数的最大公约数  gright[i]

遍历，去掉第i个数时，这最大公约数为gleft[i-1]+grigh[i+1]

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>const int maxn=100010;using namespace std;int g[maxn];int gleft[maxn];int gright[maxn];int _gcd(int a,int b){    return a==0?b:_gcd(b%a,a);}int main(){    int t,n,maxnn;     scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&g[i]);            if(i==1)                gleft[i]=g[i];            else                gleft[i]=_gcd(gleft[i-1],g[i]);        }        for(int i=n;i>0;i--)        {            if(i==n)                gright[i]=g[i];            else                gright[i]=_gcd(gright[i+1],g[i]);        }        maxnn=0;        for(int i=1;i<=n;i++)        {            if(i==1)                maxnn=max(maxnn, gright[2]);            else if(i==n)                maxnn=max(maxnn,gleft[n-1]);            else                maxnn=max(maxnn,_gcd(gleft[i-1], gright[i+1]));        }        printf("%d\n",maxnn);    }    return 0;}