Educational Codeforces Round 11--A. Co-prime Array

A. Co-prime Array

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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：给出 n 个数，要求每两个相邻的数之间两两互素（互为质数），

若两个数非互素则加入一个数使之满足条件，最后输入序列。

思路：本来是枚举两个数的公约数往上递增判断其是否分别与这两个

数互素，结果错了，后来当不满足条件时，直接往两个数之间

加入 1 即可，因为 1 与任何数互素。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
#include <utility>
#include <algorithm>
using namespace std;
#define N 1020
#define inf 0x3f3f3f3f
int num
, ans[2*N];
int gcd(int a, int b)
{
int tmp;
while (a%b){
tmp = a;
a = b;
b = tmp%b;
}
return b;
}
int main()
{
#ifdef OFFLINE
freopen("t.txt", "r", stdin);
#endif
int i, j, k, n, m, t, x;
while (~scanf("%d", &n))
{
t = x = 0;
for (i = 0; i < n; i++){
scanf("%d", &num[i]);
}
for (i = 0; i < n - 1; i++){
if (gcd(num[i], num[i + 1]) == 1){
ans[x++] = num[i];
}
else{
t++;
ans[x++] = num[i], ans[x++] = 1; //1 跟任何数都互素
}
}
ans[x++] = num[n - 1];
printf("%d\n%d", t, ans[0]);
for (i = 1; i < x; i++)
printf(" %d", ans[i]);
printf("\n");
}
return 0;
}