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

問題連結：UVA11300 Spreading the Wealth
問題描述：（略）
問題分析：
    這是藍書的一道思維題，參見藍書第一章，不解釋。
程式說明：（略）
參考連結：（略）
題記：（略）

AC的C語言程式如下：

/* UVA11300 Spreading the Wealth */

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6;
long long a[N + 1], c[N + 1], tot, m;

int main()
{
    int n;
    while(scanf("%d", &n) == 1) {
        tot = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            tot += a[i];
        }

        m = tot / n;
        c[0] = 0;
        for(int i = 1; i < n; i++) c[i] = c[i - 1] + a[i] - m;
        sort(c, c + n);
        long long x1 = c[n / 2], ans = 0;
        for(int i = 0; i < n; i++) ans += abs(x1 - c[i]);

        printf("%lld\n", ans);
    }

    return 0;
}