zoukankan      html  css  js  c++  java
  • UVA11300 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

    3
    100
    100
    100
    4
    1
    2
    5
    4
    

    Sample Output

    0
    4
    

    Problem setter: Josh Bao 

    /*
       分钱只能左右分,问最少使用的硬币的个数
       A[i]:当前拥有的钱的数目   xi:表示i分给i+1钱的数目,正表示分出,负表示收入
       M:分完之后的状态(正分出,负收入)
       x1
       M = A[1] + x1 - x2   --- x2 = A[1] - M + x1 = B[1] + x1 
       M = A[2] + x2 - x3   --- x3 = A[2] - M + x2 = B[2] + x2 = B[1] + B[2] + x1
       ....
       ....
       xn = sum(B[1] + ... + B[n-1]) + x1
       
       answer = min(abs(xi - x1) + ... + abs(xi - xn)) (1 ... n)
       中位数 
    */
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    int main()
    {
        int n;
        int a[1100000];
        int c[1100000];
        while(~scanf("%d ", &n)){
            long long  sum = 0;
            for(int i = 1; i <= n ; i++){
                scanf("%d", &a[i]);
                sum += a[i];
            }
            int temp = sum / n;
            c[0] = 0;
            for(int i = 1; i < n ; i++)
                c[i] = c[i-1] + a[i] - temp;
            sort(c , c + n);
            int x = c[n/2];
            sum = 0;
            for(int i = 0; i < n ; i++){
                sum += fabs(c[i] - x);
            }
            printf("%lld
    ", sum);
        }
        return 0;
    }
    

      

  • 相关阅读:
    在业务层进行回滚操作时如何避免回滚指令冗余
    云计算VS大数据 记与思
    [SAPBI]解决:不存在源系统(逻辑系统) T90CLNT090 的源系统标识符
    物料分类账简介
    BW Query设计中公式冲突解决方案
    解决BW处理链中节点有选择的执行
    如何立即手动执行BW周期性处理链
    资产数据源抽取当日增量数据的配置说明
    文本数据源预览出错
    主数据上载因重复记录报错问题解决
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4645576.html
Copyright © 2011-2022 走看看