设第i个人需要给第i+1个人的金币数为xi(xi为负代表收到钱),列出一大堆方程. 设第i个人给第i-1个人的钱为xi(xi<0表示第i-1个人给第i个人钱).计算出最后每个人应该有的钱m,解方程得xi=x1-(a1+a2+ … +a(i-1)-(i-1) * m)=x1-ti. 答案就是让|x1-t1|+|x1-t2|+…+|x1-tn|最小,所以x1应该取这些t中的中位数.      by sdfzyhx /* UVa11300 - Spreading the Wealth */ #incl…

题目 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…

题目大意 圆桌旁边坐着\(n\)个人,每个人有一定数量的金币,金币的总数能被\(n\)整除.每个人可以给他左右相邻的人一些金币,最终使得每个人的金币数量相等.您的任务是求出被转手的金币的数量的最小值. Solution 考虑一下这个东西怎么做,假设每一个人向周围都可以传递对吧,设\(x_i\)表示\(i\)向\(i-1\)传递的金币数量,那么\(x_1\)就是\(1\)向\(n\)传递的金币数量,那么接下来我们考虑这个东西本质是什么? 设每一个人一开始有\(a_i\)个金币,那么显然,\(a_i…

bzoj1045 洛谷P4016 洛谷P2512 bzoj3293 洛谷P3156 题解:https://www.luogu.org/blog/LittleRewriter/solution-p2512 #include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; #define fi first #define se second #…

前方数学警告 题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=25&page=show_problem&problem=2275 https://www.luogu.com.cn/problem/UVA11300 分析: 搞了这么久终于碰到一个数学题了,这也是本蒟蒻做的第一个数学题吧,值得纪念. 第一次看见这题的时候,嗯,好像是个暴力...然后就打了一个暴力,发现…

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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 th…

题目链接:uva 11300 - Spreading the Wealth 题目大意:有n个人坐在圆桌旁,每个人有一定的金币,金币的总数可以被n整除,现在每个人可以给左右的人一些金币,使得每个人手上的金币数量相等,问说最少移动的金币数额. 解题思路:假设xi为第i个人给左手边人的金币数量,那么就有a[i] - x[i]+ x[i + 1] = aver.那么 a[1] - x[1] + x[2] = aver -> x2 = aver - a[1] + x[1]  -> x[2]= x[1]…

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 th…

Spreading the Wealth Problem's Link ---------------------------------------------------------------------------- Mean: n个人围成一圈,每个人手里有Ai个金币,每个人可以给与他相邻的人一些金币,通过一系列的流转后,最后所有人的金币数相等.问整个过程最少需要流转多少金币? analyse: 这是一道很有趣的数学题. 假设有4个人,按顺序编号1,2,3,4.假设1号给2号3枚金币,2…