Greedy algorithm merges fruit

Merge fruit

This question obviously has no aftereffects for each choice.
There are several things to notice in the obvious greedy algorithm . After merging into a pile, a new pile will be formed. Two piles of merged
reordering here, according to the idea of ​​this question itself, it is easy to think of insertion sort
insertion sort:

//这种插入排序是前面已知，将手中的牌插到前面
#include<iostream>
#include<algorithm>
using namespace std;
int a[105];
int main()
{
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	//开始插入排序
	for(int i=1;i<n;i++)
	{
		for(int j=i-1;j>=0&&a[j]>a[j+1];j--)
		{
			swap(a[j],a[j+1]);
		}
	}
	for(int i=0;i<n;i++)
	{
		cout<<a[i]<<endl;
	}
	return 0; 
} 

This question obviously needs to be inserted backwards

#include<iostream>
#include<algorithm>
using namespace std;
int a[105];
int main()
{
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	int res=0;
	for(int i=1;i<n;i++)
	{
		a[i]+=a[i-1];
		res+=a[i];
		for(int j=i+1;j<n&&a[j]<a[j-1];j++)
		{
			swap(a[j],a[j-1]);
		}
	}
	cout<<res<<endl;
	return 0;
}