UVA 10137 The Trip

     每次小練習都有收穫，今天強化了一下精度控制這一塊，

 The Trip

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.

The 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.

The 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

310.0020.0030.00415.0015.013.003.010

Output for Sample Input

$10.00$11.99

                     思路很簡單 ，求出總錢數的平均值，再計算要轉移的錢數，但要注意錢數的最小單位是0.01，所以對於對平均值的處理顯得很重要，

                      sum=sum/per+0.005;這是四捨五入的方法，加0.005，若是小數點後第三位有進位 ，表明第三位>5，

                     sum=floor(sum*100)/100.0;    floor(sum*100)是捨棄原sum小數點第三位及其以後的位元，

                    還有一個要注意的地方是求轉移的錢數的時應該 有兩個變數，一個記錄比平均值大的，一個記錄比平均值小的，因為之前對平均值做了處理，所以這樣個變數未必                                 相等。

代碼：

#include<cstdio>#include<cstring>#include<cmath>using namespace std;int main(){    int per=10,i;    double num[1010],sum,a,b;    while(scanf("%d",&per)!=EOF&&per)    {        sum=0;        for(i=0; i<per; i++)        {            scanf("%lf",&num[i]);            sum+=num[i];        }        sum=sum/per+0.005;        sum=floor(sum*100)/100.0;        a=0;b=0;        for(i=0;i<per;i++)        {            if(num[i]-sum>1e-12)            a+=num[i]-sum;            else  b+=sum-num[i];        }        if(a>b)        printf("$%.2lf\n",b);        else  printf("$%.2lf\n",a);    }    return 0;}