CodeForces 660A Co-prime Array

A - Co-prime Array
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
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660A

Description

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.

Sample Input

Input

3
2 7 28

Output

1

2 7 9 28

就是判断一个数组相邻的两个数互质不互质，不互质就加个数让前后互质。样例的加的9是坑人的，因为题目最后说如果有多个答案输出任意一个就行。所以加1就行，1和任何数都互质。就是数值向后推的时候有点烦。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int GCD(int x,int y)
{
if(x%y==0)
return y;
else
return (GCD(y,x%y));
}
int a[1100],b[1100];
int main()
{
int n,i,j,cnt;
while(scanf("%d",&n)!=EOF)
{
cnt=0;
memset(a,1,sizeof(a));
memset(b,1,sizeof(b));
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
if(GCD(a[i],a[i+1])==1)
b[i+cnt]=a[i];
else if(GCD(a[i],a[i+1])!=1)
{
b[i+cnt]=a[i];
b[i+cnt+1]=1;
cnt++;
}
}
printf("%d\n",cnt);
for(i=0;i<n+cnt-1;i++)
printf("%d ",b[i]);
printf("%d\n",b[n+cnt-1]);
}
return 0;
}