数学/思维 UVA 11300 Spreading the Wealth

题目传送门

 /*
     假设x1为1号给n号的金币数(逆时针)，下面类似
     a[1] - x1 + x2 = m(平均数)  得x2 = x1 + m - a[1] = x1 - c1; //规定c1 = a[1] - m，下面类似
     a[2] - x2 + x3 = m ,x3 = m + x2 - a[2] = m + (m + x1 - a[1]) - a[2] = 2 * m + x1 - a[1] - a[2] = x1 - c2;
     a[3] - x3 + x4 = m ,x4 = m + x3 - a[3] =............................= 3 * m + x1 - a[1] - a[2] - a[3] = x1 - c3;
     而我们求得是|x1|＋|x2|+|x3|+....+|xn|
     把上边的公式带入，每一项都会变成|x1 - ci|的形式，那就变成了：在数轴上有ｎ个点，求到他们的距离和最小的点是谁？
     然后结论是x1 = ci的中位数。
     中位数证明：http://blog.csdn.net/zhengnanlee/article/details/8915098
 */
 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN], c[MAXN];

 int main(void)        //UVA 11300 Spreading the Wealth
 {
     //freopen ("UVA_11300.in", "r", stdin);

     int n;
     while (scanf ("%d", &n) == )
     {
         long long sum = ;    int ave = ;
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);
             sum += a[i];
         }
         ave = sum / n;
         c[] = ;
         for (int i=; i<=n; ++i)
         {
             c[i] = c[i-] + a[i] - ave;
         }

         sort (c+, c++n);
         int x = c[n/];    long long ans = ;
         for (int i=; i<=n; ++i)    ans += abs (x - c[i]);

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

     return ;
 }