ZOJ - 1847 The Trip（奇葩诡异坑题）

The Trip

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

Description

A number of students are members of a club that travels annually to exotic locations. Their destinations in the past have included Indianapolis, Phoenix, Nashville, Philadelphia, San Jose, and Atlanta. This spring they are planning a trip to Eindhoven.

The group agrees in advance to share expenses equally, but it is not practical to have them share every expense as it occurs. So individuals in the group pay for particular things, like meals, hotels, taxi rides, plane tickets, etc. After the trip, each student's
expenses are tallied and money is exchanged so that the net cost to each is the same, to within one cent. In the past, this money exchange has been tedious and time consuming. Your job is to compute, from a list of expenses, the minimum amount of money that
must change hands in order to equalize (within a cent) all the students' costs.

Input

Standard input will contain the information for several trips. The information for each trip consists of a line containing a positive integer, n, the number of students on the trip, followed by n lines of input, each containing the amount, in dollars and cents,
spent by a student. There are no more than 1000 students and no student spent more than $10,000.00. A single line containing 0 follows the information for the last trip.

Output

For each trip, output a line stating the total amount of money, in dollars and cents, that must be exchanged to equalize the students' costs.

Sample Input

3

10.00

20.00

30.00

4

15.00

15.01

3.00

3.01

0

Sample Output

$10.00

$11.99

Source
University of Waterloo Local Contest 1999.01.31

分析：
个人感觉这种题没什么意思，好像是出题者的坑。
浮点数的处理要熟练。
代码的处理体现了公平，没有直接取整，而是分类。
其实我也不明白为什么要这么处理，算了。
该题属于简单题，但涉及一个取整或许有所麻烦，精确解应该是求出平均值，再求偏差绝对值和除 2。这里由于精确到分，求出平均值 mean，按分求出 ceil 和 floor 的值。然后对于每个人，若钱 < mean，求与 floor 偏差绝对值加到总和中，否则与 ceil 求。另外，记录偏差(不取绝对值总和)，如果最后总偏差不为 0，要将总偏差绝对值加到总和中。最后输出总和一半即可。
ac代码：

#include <iostream>

#include<cstdio>

using namespace std;

int main()

{

int n,i;

double a[1005];//float???

double s,d;

while(scanf("%d",&n)&&n!=0)

{

s=0;

for(i=0;i<n;i++)

{

scanf("%lf",&a[i]);

s+=a[i];

//printf("a[i]=%lf,s=%lf\n",a[i],s);//

}

d=s/n;

int m=d*100;

double d2=m/100+m%100*0.01;//d2=floor(d)

double d3;

//double p1=d-d2,p2=d3-d;

double s1=0,s2=0;

//int c1=0,c2=0;

if(d==d2)

d3=d2;

else

d3=d2+0.01;//d3=ceil(d)

//printf("s=%lf,d=%lf\n",s,d);//

//double ans=0;

for(i=0;i<n;i++)

{

if(d>a[i])

{

s1+=d2-a[i];//d2,d3交换就错了,大概是出题人的习惯吧。不细究了。

//c2++;

}

else

{

s2+=a[i]-d3;

//c1++;

}

}

if(s1>s2)//取s1，s2中较大的。

printf("$%.2lf\n",s1);

else

printf("$%.2lf\n",s2);

}

return 0;

}