HDU 6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 44    Accepted Submission(s): 34

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

题意：给出一列数，去掉其中一个数，使得所有数的最大公因数最大。

枚举每一个数字，尝试去掉，然后求所有数的最大公因数，去掉就是求公因数时标记跳过。提取出前两个因子，和第三个求因子，得出的最大共因子，再和第四个数比较。。。中间适当做优化，有几种情况可以跳出循环。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;inline int GCD(int a,int b){    int t;    if(b>a)    {        t=a;        a=b;        b=t;    }    while(b)    {        t=b;        b=a%b;        a=t;    }    return a;}int main(){    int a[100001],b[100001];    int t,n,i,j;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%d",&a[i]);        }        int y,M=0;        b[0]=b[1]=-1;        for(j=0; j<n; j++)        {            y=a[0];            if(j==0)                y=a[1];            for(i=0; i<n; i++)            {                if(i==j)                    continue;                y=GCD(a[i],y);                if(y<=M)                    break;                if(i==j+1)//对比记录的共因子大小是否有变化，没变化则不用继续，节约时间                {                    if(y==b[i])                    break;                }                if(i==j+2)//记录上一组处理到这个位置的共因子大小                {                    b[i]=y;                }            }            if(M<y)                M=y;        }        printf("%d\n",M);    }    return 0;}