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

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 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;
}