uva 11300(分金币）

题目：
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

题意：
圆桌旁坐着n个人，每个人有一定数目的金币，需要将所有人的硬币进行一次平均分配，这里要注意的是，每个人只能和左右相邻的人交换硬币，求出被转手的金币数量的最小值并输出。

刘汝佳老师代码：

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 1000000 + 10;long long A[maxn], C[maxn], tot, M;int main() {  int n;  while(scanf("%d", &n) == 1) { // 输入数据大，scanf比cin快     tot = 0;    for(int i = 1; i <= n; i++) { scanf("%lld", &A[i]); tot += A[i]; } // 用%lld输入long long    M = tot / n;    C[0] = 0;     for(int i = 1; i < n; i++) C[i] = C[i-1] + A[i] - M; // 递推C数组    sort(C, C+n);    long long x1 = C[n/2], ans = 0; // 计算x1    for(int i = 0; i < n; i++) ans += abs(x1 - C[i]);     // 把x1代入，计算转手的总金币数    printf("%lld\n", ans);  }  return 0;}