Codeforces 660A Co-prime Array 【水题】

题目链接：Codeforces 660A Co-prime Array

A. Co-prime Array
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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：给定n个元素的序列，让你插入尽量少的数，使得序列相邻元素互质。

1和任何数互质。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iostream>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 1;
int a[1010];
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
        int ans = 0;
        for(int i = 0; i < n-1; i++) {
            if(gcd(a[i], a[i+1]) != 1) {
                ans++;
            }
        }
        printf("%d\n", ans);
        for(int i = 0; i < n-1; i++) {
            printf("%d ", a[i]);
            if(gcd(a[i], a[i+1]) != 1) {
                printf("%d ", 1);
            }
        }
        printf("%d\n", a[n-1]);
    }
    return 0;
}