UVa11300 Spreading the Wealth(数学问题)

题意：给出n个人，每个人有一些金币，可以给一些金币左边或者右边的人，最终使得每个人有相同的金币，问最小的转移金币是多少？

思路：可以假定给金币方向是逆时间方向，值可能是正负。M表示最终每个人有的金币，用xi表示第i个人所给的，Pi表示第i个人有的金币

有M = P1 - x1 + x2   => x2 = x1 - (P1 - M)  => x2 = x1 - c1

P2 - x2 + x3 = M => x3 = x1 - (P1 + P2 - 2M)  => x3 = x1 - c2;

Pn-1 - xn-1 + xn = M => xn = x1 - (P1+P2+...+Pn-1 - (n-1)M)  => xn = x1 - cn-1

则总的转移次数为sum = |x1|+|x1-c1|+...+|x1-cn-1|是最小值，转变为数学问题0，c1,c2,cn-1这几个点构成的数轴上，找出一个点，使得到这几个点的和最小，显然是中位点

代码如下：

#include <iostream>#include <fstream>#include <algorithm>#include <vector>using namespace std;typedef unsigned long long ULL;class SpreadingTheWealth{public:ULL spreadTheWealth(vector<ULL> v){vector<int> c(v.size());ULL sum = 0;for (size_t i = 0; i < v.size(); i++) sum += v[i];ULL m = sum / v.size();c[0] = 0;for (size_t i = 1; i < c.size(); i++){c[i] = c[i - 1] + v[i - 1] - m;}sort(c.begin(), c.end());ULL ans = 0;ULL x = c[c.size() / 2];for (size_t i = 0; i < v.size(); i++){ans += labs(c[i] - x);}return ans;}};SpreadingTheWealth solver;int main(){#ifndef ONLINE_JUDGEifstream fin("f:\\OJ\\uva_in.txt");streambuf *old = cin.rdbuf(fin.rdbuf());#endifint n;while (cin >> n){vector<ULL> v;for (int i = 0; i < n; i++){ULL coin;cin >> coin;v.push_back(coin);}cout << solver.spreadTheWealth(v) << endl;}#ifndef ONLINE_JUDGEcin.rdbuf(old);#endifreturn 0;}