Codeforces 660A：Co-prime Array（水题+思维）

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

题目大意：给你一个序列，要求你从中加一些数字，使得序列中任意两个相邻的数最大公约数为1，输出你添加的数的个数和插入数后的序列。（多种情况的话输出其中一种即可）
解题思路：当两个相邻的数最大公约数不为1时，在中间插个1即可。（.....）
代码如下：
#include <cstdio>
#include <vector>
using namespace std;
vector<int>a;
int gcd(int a,int b)
{
if(a>b)
{
int t=a;
a=b;
b=t;
}
while(b%a)
{
int t=a;
a=b%a;
b=t;
}
return a;
}
int main()
{
int n;
scanf("%d",&n);
int x;
scanf("%d",&x);
a.push_back(x);//先把第一个放进去，以后每放一个数之前与上一个数gcd一下
int num;
int cnt=0;
for(int i=1;i<n;i++)
{
scanf("%d",&x);
num=a.size();
if(gcd(x,a[num-1])!=1)
{
a.push_back(1);//插入1
cnt++;//计数器
}
a.push_back(x);//再插入x
}
int size=a.size();
printf("%d\n",cnt);
for(int i=0;i<size-1;i++)
{
printf("%d ",a[i]);
}
printf("%d\n",a[size-1]);
return 0;
}