【codeforces】-660Co-prime Array

最新推荐文章于 2016-11-01 22:55:12 发布

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本文介绍了一种名为Co-primeArray的算法问题，旨在通过最少的步骤将给定数组转换为任意两个相邻元素互质的形式。具体实现包括确定所需插入元素的数量及位置，并提供一种简单有效的解法。

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

这题我感觉还是我的英语水平太差，不然如此简单的思维题应该能想出来，一个细节没注意到，他说添加任何一个正整数，本来以为按序递增呢，好吧是我主观臆想了

直接+1........

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1010;
int a[N],b[N],vis[N];
int gcd(int a,int b) {
	if(!b)
		return a;
	else
		return gcd(b,a%b);
}
int main() {
	int n;
	while(scanf("%d",&n)!=EOF) {
		memset(vis,0,sizeof(vis));
		memset(b,0,sizeof(b));
		for(int i=0; i<n; i++) {
			scanf("%d",&a[i]);
		}
		int num=0;
		for(int i=0; i<n-1; i++) {
			if(gcd(a[i],a[i+1])==1) {
				continue;
			} else {
				b[i]=1;
				vis[i]=1;
				num++;
			}
		}
		printf("%d\n",num);
		for(int i=0; i<n; i++) {
			if(!i)
				printf("%d",a[i]);
			else
				printf(" %d",a[i]);
			if(vis[i]) {
				printf(" %d",b[i]);
			}
		}
		printf("\n");
	}
	return 0;
}