CodeForces - 660A Co-prime Array （模拟）

CodeForces - 660A

Co-prime Array

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

Submit Status

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 adding k 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

Source

Educational Codeforces Round 11

//题意：

给你一个有n个元素的序列，现在让你在序列中加入k个数（最少），使得这个序列任意的相邻的两个元素互质。输出这个最小的数k并将加入后的序列输出

//思路：

直接判断相邻两个元素是否互质（gcd(a,b)==1），不互质的话在中间加个1即可。

#include<stdio.h>#include<string.h>#include<math.h>#include<set>#include<map>#include<stack>#include<queue>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define ull unsigned long long#define ll long long#define IN __int64#define N 100010#define M 1000000007using namespace std;int a[N],b[N];int gcd(int a,int b){return b?gcd(b,a%b):a;}int main(){int n,m,i,j,k;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d",&a[i]);k=0;for(i=0;i<n;i++){if(i!=n-1){if(gcd(a[i],a[i+1])==1)b[k++]=a[i];elseb[k++]=a[i],b[k++]=1;}}b[k++]=a[n-1];printf("%d\n",k-n);for(i=0;i<k;i++)printf("%d ",b[i]);printf("\n");}return 0;}