Uva 11300 Spreading the Wealth (代数分析 + 中位数)

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#中位数 #数学分析

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本文介绍了一个共产主义政权尝试在一个村庄中平均分配财富的问题。通过将个人财产转化为等值硬币并确保总数能被村民数量整除，算法计算了每个人向左右邻居转移硬币的最小数量，从而实现财富平等分配。

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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.
Input
There is a number of inputs. Each input begins with n (n < 1000001), the number of people in the
village. n lines 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.
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
PS:lrj小白书p4上面的题。这题难度算比较大的。

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e6+4;
long long a[maxn];
long long c[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    long long ave, sum;
    while(cin>>n)
    {
        sum = 0;
        for(int i = 1; i <= n; i++)
        {
            cin>>a[i];
            sum += a[i];
        }
        ave = sum / n;
        c[0] = 0;
        for(int i = 1; i < n; i++)c[i] = c[i-1] + a[i] - ave;
        sort(c, c+n);
        long long x1 = c[n/2], ans = 0;
        for(int i = 0; i < n; i++)ans += abs(x1 - c[i]);
        cout<<ans<<endl;
    }
    return 0;
}