CodeForces 660 A. Co-prime Array【互质+思维】

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本文介绍了一道名为Co-primeArray的编程题，任务是在给定数组中插入尽可能少的正整数，使相邻元素互质。文章提供了题目的详细说明及一个简单有效的解法，利用1作为通用互质数。

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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 adding k 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

题意：

给出一个序列，问不改变原先序列的顺序，需要加入多少个数，使得得到的新序列相邻的两个数字互质，并且输出满足要求的序列

题解：

首先感觉如果相邻两个数字不互质，想找到一个和他们俩都互质的数好难，后来发现，自己想多了...找1 就可以了....

然后果断提交....wa....才发现忘了统计需要加入的个数....

/*
http://blog.youkuaiyun.com/liuke19950717
*/
#include<cstdio>
int gcd(int a,int b)
{
	if(!b)
	{
		return a;
	}
	return gcd(b,a%b);
}
int co_prime(int a,int b)
{
	return gcd(a,b)==1;
}
int main()
{
	//freopen("shuju.txt","r",stdin);
	int n,array[1005]={0};
	while(~scanf("%d",&n))
	{
		int cnt=0,vis[1005]={0};//不想再次判断相邻互质，加上标记就行
		for(int i=0;i<n;++i)
		{
			scanf("%d",&array[i]);
			if(i&&!co_prime(array[i-1],array[i]))
			{
				++cnt;vis[i]=1;
			}
		}
		int i=0;
		printf("%d\n%d",cnt,array[i++]);
		while(i<n)
		{
			if(vis[i])
			{
				printf(" 1");
			}
			printf(" %d",array[i++]);
		}
		printf("\n");
	}
	return 0;
}