Educational Codeforces Round 11——A. Co-prime Array（map+vector）-优快云博客

本文探讨了在不改变序列相对顺序的前提下，如何通过最少次数的插入操作使序列中任意两个相邻数互质的方法。介绍了具体的实现思路及算法步骤，并提供了完整的示例代码。

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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 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 kelements 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

给你一个序列在不改变原来相对顺序的情况下插入最少个数使得该序列相邻数均为互质，只要插入的个数最少，答案可以不唯一。本来以为会WA，没想到过了。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
inline int gcd(const int &a,const int &b)
{
	return b?gcd(b,a%b):a;
}
int main (void)
{
	int n,i,j,ans,t;	
	while (cin>>n)
	{
		vector<int>vec;
		map<int,int>pos;
		for (i=0; i<n; i++)
		{
			cin>>t;
			vec.push_back(t);			
		}
		for (i=0; i<n-1; i++)
		{
			if(gcd(vec[i],vec[i+1])!=1)
			{
				for (j=2; j<=100000000; j++)
				{
					if(gcd(vec[i],j)==1&&gcd(vec[i+1],j)==1)//map记录每个位置是否要放入和放入的值
					{
						pos[i]=j;
						break;
					}
				}
			}
		}
		cout<<pos.size()<<endl;
		for (i=0; i<n; i++)
		{
			if(pos.find(i)!=pos.end())
				cout<<vec[i]<<" "<<pos[i];
			else 
				cout<<vec[i];
			if(i==n-1)
				cout<<endl;
			else
				cout<<" ";
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5380418.html