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. 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.
The Output
For each input, output the minimum number of coins that must be transferred on a single line.
Sample Input
代码语言：javascript复制3
100
100
100
4
1
2
5
4Sample Output
代码语言：javascript复制0
4题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2275
题意：圆桌坐着N个人，每个人有一定的金币，金币总数能被N整除。每个人能给左右相邻的人一些金币，最终使得每个人的金币数目相等，求被转手金币数量的最小值。
设 xi表示i号给i-1号xi金币，若xi为负，这表示i-1号给i号(-xi)个金币
Ai表示i号一开始持有的金币
则：对与第1个人:A1-X1 X2=M ===>X2=X1-(A1-M);令C1=A1-M
对于第2个人：A2-X2 X3=M ====>x3=x2-(A2-M) ====>x3=x1-(A1 A2-2M)；===>x3=x1-C2;
……
对于第n个人：An-Xn x1=M 这个是个恒等式，无用；
所以我们的答案应该是 |X1| |X2| |X3| …… |Xn|的最小值；
====>|X1| |X1-C1| |X1-C2| …… |Xn-1-Cn|的最小值
故当X1取C数组的中间值时，结果最小。。。
注意：|X1 – Ci|在数轴上就是x1到Ci的距离，所以问题变成了：给定数轴上的n个点，找出一个到它们的距离之和尽量小的点。
这个最优的X1就是这些数的“中位数”。即排序以后位于中间的数。至于证明大家自己搜索吧~
下面给出AC代码：
代码语言：javascript复制 1 #include <bits/stdc .h>
2 using namespace std;
3 const int maxn=1000005;
4 long long A[maxn],C[maxn],tot,M;
5 int main()
6 {
7 int n;
8 while(scanf("%d",&n)==1)//这里写EOF会超时，我也不知道咋回事
9 {
10 tot=0;
11 for(int i=1;i<=n;i )
12 {
13 scanf("%lld",&A[i]);
14 tot =A[i];
15 }
16 M=tot/n;//求平均值
17 C[0]=0;
18 for(int i=1;i<n;i )
19 C[i]=C[i-1] A[i]-M;//有个递推公式，数学的博大精深啊！
20 sort(C,C n);
21 long long x1=C[n/2];
22 long long ans=0;
23 for(int i=0;i<n;i )
24 ans =abs(x1-C[i]);
25 printf("%lldn",ans);
26 }
27 return 0;
28 }
