版权声明:转载请注明出处。 https://blog.csdn.net/u014427196/article/details/46383439

Description

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

代码:

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <string>  
#include <queue>  

using namespace std;

int n;
long long p[1010000];
long long c[1010000];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        long long sum = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%lld", &p[i]);
            sum += p[i];
        }
        long long t = sum / n;
        c[0] = 0;
        for (int i = 1; i < n; i++) c[i] = c[i - 1] + p[i] - t;
        sort(c,c+n);
        long long tmp = c[n / 2], ans = 0;
        for (int i = 0; i < n; i++) ans += abs(tmp - c[i]);
        printf("%lld\n",ans);
    }
    return 0;
}