1301. The Trip
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
$11.99
Source: Waterloo
Local Contest Jan. 31, 1999
几点注意:
1、
2、
3、
4、
/*
#include <stdio.h>
#include <string.h> float cost[100];
int cost100[100]; void sort(int num); int main(void)
{
int num,i,sum=0;
int avg,last;
int change_sum=0;
double change; scanf("%d",&num);
while(num != 0){
for(i=0;i<num;i++){
scanf("%f",&cost[i]);
cost100[i] = 100*cost[i] + 0.5;
sum += cost100[i];
}
avg = sum/num;
last = sum%num;
sort(num);
for(i=0;i<num;i++){
if(i<last)
cost100[i] = cost100[i] - avg - 1;
else
cost100[i] = cost100[i] - avg;
if(cost100[i] > 0)
change_sum += cost100[i];
} change = ((double)change_sum)/100;
printf("$%.2f\n",change); memset(cost,0,num*sizeof(float));
memset(cost100,0,num*sizeof(int)); sum = 0;
change_sum = 0; scanf("%d",&num);
} //system("PAUSE");
return 0;
}
void sort(int num)
{
int i,j;
for(i=0;i<num;i++){
for(j=0;j<num;j++){
if(cost100[j]<cost100[j+1]){
int temp = cost100[j];
cost100[j] = cost100[j+1];
cost100[j+1] = temp;
}
}
} }
*/ #include <stdio.h>
#define num 1005
double s[num];
int main()
{
int n, t, i;
while (scanf("%d", &n) == && n)
{
double sum = , resultH = , resultL = ;
for (i = ; i < n; i++)
{
scanf("%lf", &s[i]);
sum += s[i];
}
sum /= n; for (i = ; i < n; i++)
if (s[i] < sum)
{
resultL += (int)((sum - s[i])*) / 100.0;
}
else
resultH += (int)((s[i] - sum)*) / 100.0;
printf("$%0.2lf\n", resultL < resultH ? resultH:resultL);
}
return ;
}
1301. The Trip的更多相关文章
- Lesson 4 An existing trip
Text I have just received a letter from my brother,Tim. He is in Australia. He has been there for si ...
- dp or 贪心 --- hdu : Road Trip
Road Trip Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 29 ...
- 【poj1041】 John's trip
http://poj.org/problem?id=1041 (题目链接) 题意 给出一张无向图,求字典序最小欧拉回路. Solution 这鬼畜的输入是什么心态啊mdzz,这里用vector储存边, ...
- 三分 --- POJ 3301 Texas Trip
Texas Trip Problem's Link: http://poj.org/problem?id=3301 Mean: 给定n(n <= 30)个点,求出包含这些点的面积最小的正方形 ...
- HDOJ 1301
9852303 2013-12-18 11:47:01 Accepted 1301 0MS 264K 1117 B C++ 泽泽 Jungle Roads Time Limit: 2000/1000 ...
- 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem C: The Trip(水题)
Problem C: The Trip Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 19 Solved: 3[Submit][Status][Web ...
- hdu 3018 Ant Trip 欧拉回路+并查集
Ant Trip Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem ...
- URAL 1004 Sightseeing Trip(最小环)
Sightseeing Trip Time limit: 0.5 secondMemory limit: 64 MB There is a travel agency in Adelton town ...
- Codeforces Round #365 (Div. 2) Mishka and trip
Mishka and trip 题意: 有n个城市,第i个城市与第i+1个城市相连,他们边的权值等于i的美丽度*i+1的美丽度,有k个首都城市,一个首都城市与每个城市都相连,求所有边的权值. 题解: ...
随机推荐
- devices-list
转自:https://www.kernel.org/pub/linux/docs/lanana/device-list/devices-2.6.txt LINUX ALLOCATED DEVICES ...
- Android android:gravity属性介绍及效果图
转自: http://blog.csdn.net/aminfo/article/details/7784229 Android:gravity的属性官方说明如下: public static fina ...
- ASP.NET MVC 使用带有短横线的html Attributes(转载)
转载地址:http://www.nmtree.net/2013/10/25/asp-net-mvc-use-dash-in-html-attributes.html 情景再现 我们常常需要一个文本框来 ...
- 图结构练习——最小生成树(prim算法(普里姆))
图结构练习——最小生成树 Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 有n个城市,其中有些城市之间可以修建公路,修建不同 ...
- 【openGL】指定着色模型
#include "stdafx.h" #include <GL/glut.h> #include <stdlib.h> #include <math ...
- wp8 入门到精通 Utilities类 本地存储+异步
public class CCSetting { public async static void AddOrUpdateValue<T>(string key, T value) { t ...
- 发送http请求get方法
//获取网页html NSURL* url = [NSURL URLWithString:@"http://www.baidu.com"]; NSMutableURLRequest ...
- map[C++]
//map是一个存储键值对的容器,也是一个双向链表 #include <iostream> using namespace std; #include <map> int ma ...
- 遍历Map
Map map = new HashMap(); map.put("1", "value1"); map.put("2", "va ...
- java测试时常见的一些错误
1.解决警告: [SetPropertiesRule]{Server/Service/Engine/Host/Context} Setting property 'source' http://blo ...