分情况讨论，双指针移动（附代码）

```
#include "pch.h"
#include <iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n, sum = 0,count=0, num[105] = {0}; cin >> n;
	int tail = n, head = 1;
	for (int i = 1; i <= n; ++i)
	{
		cin >> num[i];
		sum += num[i];
	}
	sum /= n;
	while (head < tail)
	{
		while (num[head] == sum&& head < tail)head++;//等于的时候直接移动指针
		while (num[tail] == sum && head < tail)tail--;//移动指针
		while (num[head] > sum&& head < tail)
		{
			num[head + 1] += num[head] - sum;
			num[head] = sum;
			head++;
			count++;
		}
		while (num[tail] > sum&& head < tail)//大于的时候移动指针并且加1
		{
			num[tail - 1] += num[tail] - sum;
			num[tail] = sum;
			tail--;
			count++;
		}
		int i = 0, j = 0,left=0,right=0;
		for (i = 0;; ++i)//两边都小于的时候选取需要交换次数最少的完成操作
		{
			left += num[i + head];
			if (left >= sum)break;
		}
		for (j = 0;;++j)
		{
			right += num[tail - j];
			if (right >= sum)break;
		}
		if (i < j)
		{
			left -= num[head + i];
			count += i ;
			for (int k = 1; k < i; ++k)num[k+head] = 0;//过程中的都被赋值为0
			num[head+i] -= (sum - left);
			head++;
		}
		else
		{
			count += j;
			right -= num[tail - j];
			for (int k = 1; k < j; ++k)num[tail - k] = 0;
			num[tail-j] -= (sum - right);
			tail -- ;
		}
	}
	cout << count;
	return 0;
}