UVa 11300 Spreading the Wealth (使用方程进行等价转化&快速选择优化)

11300 - Spreading the Wealth

Time limit: 6.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=456&page=show_problem&problem=2275

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. nlines 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

《训练指南》P5-6上的一道好题。

可以用快速选择算法实现。（加上快速io后，本文发表时排在第2名）

排序：

/*0.242s*/

#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

long long x[1000010];

int main(void)
{
	long long n, i, ave, d, ans;
	while (~scanf("%lld", &n))
	{
		ave = 0;
		for (i = 0; i < n; i++)
		{
			scanf("%lld", &x[i]);
			ave += x[i];
		}
		ave /= n;
		x[0] = 0;
		for (i = 1; i < n; i++)
			x[i] += x[i - 1] - ave;
		sort(x, x + n);
		d = x[n / 2];//取中位数
		ans = 0;
		for (i = 0; i < n; i++)
			ans += abs(d - x[i]);
		printf("%lld\n", ans);
	}
	return 0;
}

快速选择：

/*0.185s*/

#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

long long x[1000010];

int main(void)
{
	long long n, i, ave, d, ans;
	while (~scanf("%lld", &n))
	{
		ave = 0;
		for (i = 0; i < n; i++)
		{
			scanf("%lld", &x[i]);
			ave += x[i];
		}
		ave /= n;
		x[0] = 0;
		for (i = 1; i < n; i++)
			x[i] += x[i - 1] - ave;
		nth_element(x, x + (n >> 1), x + n);
		d = x[n >> 1];///取中位数
		ans = 0;
		for (i = 0; i < n; i++)
			ans += abs(d - x[i]);
		printf("%lld\n", ans);
	}
	return 0;
}