A problem of arranging the elements in an array in ascending order in language?

Summary. You can put the elements of both arrays into one array and then use the sort() function.

C Array combines an ascending array and an array in descending order into one of the arrays.

You can put the elements of both arrays into one array and then use the sort() function.

You can follow your steps.

8. i=9;

9. a[i] >x

10. a[i+1] = x

For 8, the first copy needs to find the end of the array, so that bai can find it from back to front. For 9, we need to find the position where the x input zhi is greater than the number of dao in the array, so if the current array element is greater than x, we need to look forward.

For 10, fill in the program according to the comments.

Here are the results:

This kind of operation of arrays is difficult and inefficient, and this situation is now handled by containers in STL (standard template library): vector, deque, list, map, etc.

It is better to have the original question.

The array you said above is 12345, enter 6, naturally it is in the last place, this is very easy to implement but if you enter 3, does this mean that this array does not change, or 12345?

If you enter 7, 123457?

But if the original data is not regular.

How to design the original rules?

#include

Sort function. void sort(int a,int n)if(max!=i)}}

int main()

sort(a,10);

a is an array that needs to be sorted, and there are 10 of them.

for(i=0;i<10;i++)

printf("%d, ",a[i]);The case before insertion printf("");

scanf("%d",&a[10]);

sort(a,11);

for(i=0;i<11;i++)

printf("%d, ",a[i]);The case after inserting printf("");

return 0;}

The loop of inserting x is followed by a failed insertion, i.e. x is placed at the end of a new array (length n+1); Then try to insert y into this new array, and the number of loops is one more than the one on the shouting field (iImplementation**:

#include

using namespace std;

int main()

int n, i, j;

int a[20];

int x, y;

cin >>n;

for (i = 0; i < n; i++)

cin >>a[i];

cin >>x >>y;

for (i = 0; i < n;i++) try to insert x into the original array of silver infiltration of length n.

if (a[i] >x)

for (j = n; j > i; j--)

a[j] =a[j - 1];

a[i] =x;

break;

if (i ==n) puts it at the end a[n] if x can't insert an array.

a[i] =x;

for (i = 0; i < n + 1;i++) attempts to insert y into an array of length n+1 with x.

if (a[i] >y)

for (j = n + 1; j > i; j--)

a[j] =a[j - 1];

a[i] =y;

break;

if (i ==n+1) puts it at the end a[n+1] if y can't insert an array.

a[i] =y;

for (i = 0; i < n + 2; i++)

cout