数学/思维 UVA 11300 Spreading the Wealth

 1 /*
 2     假设x1为1号给n号的金币数(逆时针)，下面类似
 3     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
 4     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;
 5     a[3] - x3 + x4 = m ,x4 = m + x3 - a[3] =............................= 3 * m + x1 - a[1] - a[2] - a[3] = x1 - c3;
 6     而我们求得是｜ｘ１｜＋｜ｘ２｜＋｜ｘ３｜＋。。。。＋｜ｘｎ｜
 7     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？
 8     然后结论是ｘ１＝Ｃｉ的中位数。
 9     中位数证明：http://blog.csdn.net/zhengnanlee/article/details/8915098
10 */
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <cstring>
15 #include <string>
16 #include <cmath>
17 using namespace std;
18 
19 const int MAXN = 1e6 + 10;
20 const int INF = 0x3f3f3f3f;
21 int a[MAXN], c[MAXN];
22 
23 int main(void)        //UVA 11300 Spreading the Wealth
24 {
25     //freopen ("UVA_11300.in", "r", stdin);
26 
27     int n;
28     while (scanf ("%d", &n) == 1)
29     {
30         long long sum = 0;    int ave = 0;
31         for (int i=1; i<=n; ++i)
32         {
33             scanf ("%d", &a[i]);
34             sum += a[i];
35         }
36         ave = sum / n;
37         c[0] = 0;
38         for (int i=1; i<=n; ++i)
39         {
40             c[i] = c[i-1] + a[i] - ave;
41         }
42 
43         sort (c+1, c+1+n);
44         int x = c[n/2];    long long ans = 0;
45         for (int i=1; i<=n; ++i)    ans += abs (x - c[i]);
46 
47         printf ("%lld\n", ans);
48     }
49 
50     return 0;
51 }