UVA 11300 - Spreading the Wealth(数学推导+中位数)

 F. 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

310010010041254

Sample Output

04

题意:n个人坐成一圈，每个人有一些钱，现在要平分这些钱，每个人只能把钱给周围的人，问最少要转移多少钱才能平分。

思路：推导，每个人最终得钱数可以算出为M，对于第i个人来说，Xi为他给上一个人的钱，如此一来 X(i+1) = M - Ai + Xi；

X2 = M - A1 + X1 = X1 - C1.依次类推，Xi + 1 = X1 - Ci. Ci数组是可以递推出来的，然后答案就是X1 + |X1 - C1| + |X1 - C2| 。。。。 + |X1 - Cn -1|。中位数为最佳答案。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 1000005;int n;long long A[N], C[N], sum, M, ans;int main() {    while (~scanf("%d", &n)) {sum = ans = 0;for (int i = 1; i <= n; i ++) {    scanf("%lld", &A[i]);    sum += A[i];}M = sum / n;for (int i = 1; i < n ; i ++)    C[i] = C[i - 1] + A[i] - M;sort(C, C + n);int x = C[n / 2];for (int i = 0; i < n; i ++)    ans += abs(C[i] - x);printf("%lld\n", ans);    }    return 0;}