分析：
把每一個人的個數表示出來，如第一個人：A1 - X1 + X2 = M
可得X2 =M - A1 + X1 = X1 - C1(令C1 = M - A1)
以此類推。
最後找到規律，轉化為數軸上一個點到N個點之間距離的問題。
發現當x取得c的中位數時最小，累加距離得出答案。

#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <deque>
#include <list>
 

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <bitset>
#include <string>
#include <numeric>
#include <algorithm>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstring>
#include <cmath>
 

#include <cstdlib>
#include <cctype>
#include <complex>
#include <ctime>

typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

LL a[1000005];
LL c[1000005];

int main()
{
    //freopen("int.txt","r",stdin);
    //freopen("out.txt","w",stdout);
 

    int N;
    while(scanf("%d",&N) == 1)
    {
        LL sum = 0;
        for(int i = 1;i <= N;i++)
        {
            scanf("%I64d",&a[i]);
            sum += a[i];
        }
        LL M = sum / N;
        c[0] = 0;
        for(int i = 1;i < N;i++)
            c[i] = c[i - 1] + a[i] - M;
        sort(c,c + N);
        LL x1 = c[N / 2];
        LL ans = 0;
        for(int i = 0;i < N;i++)
            ans += abs(x1 - c[i]);
        printf("%I64d\n",ans);
    }
    return 0;
}

//a數組事實上能夠不要
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <bitset>
#include <string>
#include <numeric>
#include <algorithm>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <complex>
#include <ctime>


typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

LL c[1000005];

int main()
{
    //freopen("int.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int N;
    LL M,sum;
    while(scanf("%d",&N) == 1)
    {
        sum = 0;
        LL a;
        c[0] = 0;
        for(int i = 1;i <= N;i++)
        {
            scanf("%lld",&a);
            sum += a;
            c[i] = sum;
        }
        M = sum / N;
        for(int i = 1;i < N;i++)
            c[i] -= M * i;
        sort(c,c + N);
        LL x1 = c[N / 2],ans = 0;
        for(int i = 0;i < N;i++)
            ans += abs(x1 - c[i]);
        printf("%lld\n",ans);
    }
    return 0;
}

UVa 11300 - Spreading the Wealth