7-6 simplified insertion sort (15 minutes)

7-6 simplified insertion sort (15 points)
This problem requires programming, a given integer into the originally ordered sequence of integers, so that the results are still ordered sequence.

Input format:
Enter the first line gives the first non-negative integer N (<10); a second line gives the N integers from small to large in good order; and a third row gives the integer X.

Output format:
the output will remain X inserted ascending ordered sequence of integers in a row, behind each number has a space.

Sample input:
. 5
. 1 2. 4. 5. 7
. 3
Output Sample:
123457

#include<iostream>
using namespace std;

void input(int a[],int n);

int main()
{
    int n,i,j,x,a[11];

    cin>>n;
    input(a,n);                 //输入n个数到数组a中
    cin>>x;                     //输入插入的数 
   i=0;
	while(a[i]<x && i<n)
         i++;                 //右移
	for (j=n-1;j>=i;j--)
	{
		a[j+1]=a[j];          //左移
	}
	a[i]=x;                   //判断插入的数在哪位
	for(i=0;i<n+1;i++)
	cout<<a[i]<<" ";
    return 0;
}
void input(int a[],int n)
{
    for(int i=0;i<n;i++)
    cin>>a[i];
}