CodeForces 660A Co-prime Array（GCD）

A. Co-prime Array

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given an array of n elements, you must make it a co-prime array in as few moves as possible.

In each move you can insert any positive integral number you want not greater than 109 in
any place in the array.

An array is co-prime if any two adjacent numbers of it are co-prime.

In the number theory, two integers a and b are
said to be co-prime if the only positive integer that divides both of them is 1.

Input

The first line contains integer n (1 ≤ n ≤ 1000)
— the number of elements in the given array.

The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of the array a.

Output

Print integer k on the first line — the least number of elements needed to add to the array a to
make it co-prime.

The second line should contain n + k integers aj —
the elements of the array a after adding k elements
to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by
addingk elements to it.

If there are multiple answers you can print any one of them.

Example

input

3
2 7 28

output

1
2 7 9 28

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题意：给你n个数，如果相邻两数之间的最大公约数不是1，那么就在其中加一个数，使他们的最大公约数为1，增加的数只要使他们的最大公约数为1，就满足条件。

可以用vector容器来做，判断最大公约数是否为1，如果不为1，就添加个1

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>vec;
int GCD(int a,int b)
{
if(a%b==0)
return b;
else return GCD(b,a%b);
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int a;
scanf("%d",&a);
vec.push_back(a);
for(int i=1;i<n;i++)
{
scanf("%d",&a);
int u=vec.size();
if(GCD(a,vec[u-1])!=1)
vec.push_back(1);
vec.push_back(a);
}
int i;
int ans=vec.size()-n;
printf("%d\n",ans);
for(i=0;i<vec.size()-1;i++)
printf("%d ",vec[i]);
printf("%d\n",vec[i]);
}
return 0;
}