CodeForces 660A Co-prime Array（互质数列）

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本文解析了CodeForces上的一道A级题目——Co-prime Array，介绍了解题思路和核心代码实现。通过在非互质的相邻元素间插入数字1的方法，使数组满足任意相邻元素互质的要求。

题目：http://codeforces.com/problemset/problem/660/A

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

题意：

给定的序列变成任意两个相邻间时互质（最小公倍数是 1）的。

思路：

注意到给定的 The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

我们可以把两个相邻的不是互质数字中间放个 1 就行了。

Code：

#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;
const int MYDD=1103;

int GCD(int x,int y) {
	if(y==0)	return x;
	return GCD(y,x%y);
}

int main() {
	int n,a[MYDD];
	int Ans[MYDD*2];
	scanf("%d",&n);
	for(int j=0; j<n; j++)
		scanf("%d",&a[j]);
	int v=0;
	for(int j=0; j<n-1; j++ ) {
		Ans[v]=a[j];
		if(GCD(a[j],a[j+1])!=1) {/*判断互质*/
			Ans[v+1]=1;
			v=v+1;
		}
		v++;
	}
	Ans[v++]=a[n-1];
	printf("%d\n",v-n);
	for(int j=0; j<v; j++)
		printf("%d ",Ans[j]);
	return 0;
}
/*
4
2 8 5 7
6
2 6 9 8 7 15
*/