题目链接： Spreading the Wealth
Problem
A Communist regime is trying to redistribute wealth in a village. They have have decided to sit everyone around a circular table. First, everyone has converted all of their properties to coins of equal value, such that the total number of coins is divisible by the number of people in the village. Finally, each person gives a number of coins to the person on his right and a number coins to the person on his left, such that in the end, everyone has the same number of coins. Given the number of coins of each person, compute the minimum number of coins that must be transferred using this method so that everyone has the same number of coins.
The Input
There is a number of inputs. Each input begins with n(n<1000001), the number of people in the village. nlines follow, giving the number of coins of each person in the village, in counterclockwise order around the table. The total number of coins will fit inside an unsigned 64 bit integer.
The Output
For each input, output the minimum number of coins that must be transferred on a single line.
Sample Input
3
100
100
100
4
1
2
5
4
Sample Output
0
4
解题思路：这个题目一看到题目感觉不是很复杂，觉得其中应该有一定的数学关系，但是推了一会并没有得出答案，只想到了一个距离的关系。书上这个题目给出的方法十分的好，之前也没有见过这个方法。
假设对所有的人进行从1到n进行编号，第1个人给第n个人x1枚金币，给第2个人x2金币，第二个人给第一个人-x2枚金币，给第三个人x3枚金币……..
从这个方面我们可以得到一个公式
A1 - x1 + x2 = M -> x2 = M - A1 + x2 = x1 - C1(规定C1 = A1 - M）
A2 - x2 + x3 = M -> x3 = M - A2 + x2 = 2M - A1 - A2 + x1 = x1 - C2
……
所有的交换金币次数为x1到xn的绝对值的和
找到中位数即可求的最小值
代码如下：
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll arr[1000000],brr[1000000];
int main()
{
ll n;
while (cin >> n)
{
ll sum = 0;
for (int i = 1; i <= n; ++i)
{
cin >> arr[i];
sum += arr[i];
}
ll num = sum / n;
brr[0] = 0;
for (int i = 1; i < n; ++i)
{
brr[i] = brr[i - 1] + arr[i] - num;
}
sort(brr, brr + n);
ll x1 = brr[n / 2];
ll ans = 0;
for (int i = 0; i < n; ++i)
ans += abs(x1 - brr[i]);
cout << ans << endl;
}
return 0;
}