CodeForces 660A Co-prime Array (互质队列)

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

判断相邻两个数是否互质（即是否公约数为1），1和任何数都互质，所以如果相邻两个数不互质，就添加1；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1100];
int b[1100];
int gcd(int x,int y)
{
if(y==0)
return x;
else
return gcd(y,x%y);
}
int main()
{
int n,i,j;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int k=0;
for(i=0;i<n-1;i++)
{
if(gcd(a[i],a[i+1])==1)
{
b[k++]=a[i];
}
else
{
b[k++]=a[i];
b[k++]=1;
}
}
b[k++]=a[n-1];
printf("%d\n",k-n);
for(i=0;i<k-1;i++)
{
printf("%d ",b[i]);
}
printf("%d\n",b[k-1]);
return 0;
}