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


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each test case, print a single line containing a single integer, denoting the maximum GCD.
 

[align=left]Sample Input[/align]

3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8

 
[align=left]Sample Output[/align]

1
2
2

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;
}