CF660A -- Co-prime Array

Co-prime Array

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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 a_i (1 ≤ a_i ≤ 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 a_j — 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

32 7 28

Output

12 7 9 28

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题目大意：给出一个含N个元素的数列，在数列个元素之间可以随意插入任何小于10^9的正整数，每插入一个数算一步，求令这个数列相邻元素互质的最少的操作步数。

好吧，先来看一组奇葩的数据：

1001 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1

Output

191 1 1 1 2 1 1 1 1 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1

刚开始理解的是要大于前一项小于后一项，判断了半天，结果超时。。其实这道题还是So Easy的，毕竟是CF的A题。。既然要让这个数列的元素满足两两互质，那么直接在不互质的两个元素之间插入1不就OK了？！！白白浪费那么多时间。。

代码如下：

#include<cstdio>  #include<cstring>#include<algorithm>  using namespace std; int a[1010],b[1010];int gcd(int x,int y){if(y==0)return x;return gcd(y,x%y);}//求最大公约数 int main(){int n;while(~scanf("%d",&n)){for(int i=0;i<n;i++)scanf("%d",&a[i]);int ans=0;//用来记录步数 int k=0;//记录数列b的个数 for(int i=0;i<n-1;i++){if(gcd(a[i],a[i+1])==1)b[k++]=a[i];//如果互质则直接将该元素赋值给数列b else{ans++;b[k++]=a[i];//不互质的两个数中第一个数仍赋值给b b[k++]=1;//在不互质的第一个数后插入1 }//不互质的第二个数在下一步操作中处理 }b[k++]=a[n-1];//最后一个数，在循环中没考虑到！！ printf("%d\n%d",ans,b[0]);for(int i=1;i<n+ans;i++){printf(" %d",b[i]);//输出格式 }printf("\n");}    return 0;  }