ACM: 数学题 uva 11300

F. Spreading theWealth

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 ofequal value, such that the total number of coins is divisible bythe number of people in the village. Finally, each person gives anumber of coins to the person on his right and a number coins tothe person on his left, such that in the end, everyone has the samenumber of coins. Given the number of coins of each person, computethe minimum number of coins that must be transferred using thismethod so that everyone has the same number of coins.
The Input

There is a number of inputs. Each input begins withn(n<1000001), the number of people in the village. nlines follow, giving the number of coins of each person in thevillage, in counterclockwise order around the table. The totalnumber of coins will fit inside an unsigned 64 bit integer.
The Output

For each input, output the minimum number of coins that must betransferred on a single line.
Sample Input

3
100
100
100
4
1
2
5
4

Sample Output

0
4

题意: n个人围一圈, 每个人又a[i]个金币, 现在要用最少的转移金币数量使得每个人得金币相同.

解题思路:
     1. 假设第1个人分配给第n个人得金币是x1, 第二个人分配给第一个人得金币x2, 以此类推.
        原有金币a[i], 即第1个人最终金币数量是a[1]-x1+x2, 设aver为平均值, 得到以下公式:
        a[1]-x1+x2 = aver;
        a[2]-x2+x3 = aver;
        a[3]-x3+x4 = aver;
          ......
        a[n-1]-xn-1+xn = aver;
        a[n]-xn+x1 = aver;
     2. 那么每个人的变化值是x1,x2,x3,...,xn; 由上面等式可得:
        x1 = x1;
        x2 = x1-(a[1]-aver);
        x3 = x1-( (a[1]-aver)+(a[2]-aver) );
        x4 = x1-( (a[1]-aver)+(a[2]-aver)+(a[3]-aver) );
           ......
        xn-1 = x1-( (a[1]-aver)+(a[2]-aver)+(a[3]-aver)+...+(a[n-2]-aver));
        xn = x1-((a[1]-aver)+(a[2]-aver)+(a[3]-aver)+...+(a[n-1]-aver));
        设上面后缀为Ci; C1 = (a[1]-aver), C2 = ((a[1]-aver)+(a[2]-aver));以此类推,得:
        x1 = x1;
        x2 = x1-C1;
        x3 = x1-C2;
           ......
        xn = x1-Cn-1;
     3. 由上面1,2的推导结果显然是result =min(|x1|+|x1-C1|+|x1-C2|+...+|x1-Cn-1|);
        要使得result结果最小显然只跟x1的选择有关, |x1-Ci|可以看成是x1到Ci的距离(坐标轴上).
        当x1是Ci中位数时, 结果显然是最优解.

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAX 1000005

int n;
long long c[MAX], a[MAX];

inline long long my_abs(long long a)
{
    return a< 0 ? (-a) : a;
}

int main()
{
    int i;
//   freopen("input.txt", "r", stdin);
   while(scanf("%d", &n) != EOF)
    {
       long longsum = 0;
       for(i = 1; i<= n; ++i)
       {
         scanf("%lld", &a[i]);
          sum +=a[i];
       }
       long longaver = sum/n;
       c[0] =0;
       for(i = 1; i< n; ++i)
          c[i] =c[i-1]+a[i]-aver;
       sort(c,c+n);
       long longresult = 0;
       for(i = 0; i< n; ++i)
          result +=my_abs(c[n/2]-c[i]);
      printf("%lld\n", result);
    }

    return0;
}