USACO1.1.2 Greedy Gift Givers
Greedy Gift Givers
A group of NP (2 ≤ NP ≤ 10) uniquely named friends hasdecided to exchange gifts of money. Each of these friends might ormight not give some money to any or all of the other friends.Likewise, each friend might or might not receive money from any orall of the other friends. Your goal in this problem is to deduce how much more money each person gives than they receive.
The rules for gift-giving are potentially different than youmight expect. Each person sets aside a certain amount of money togive and divides this money evenly among all those to whom he orshe is giving a gift. No fractional money is available, so dividing3 among 2 friends would be 1 each for the friends with 1 left over-- that 1 left over stays in the giver's "account".
In any group of friends, some people are more giving than others(or at least may have more acquaintances) and some people have moremoney than others.
Given a group of friends, no one of whom has a name longer than14 characters, the money each person in the group spends on gifts,and a (sub)list of friends to whom each person gives gifts, determinehow much more (or less) each person in the group gives than theyreceive.
IMPORTANT NOTE
The grader machine is a Linux machine that uses standard Unixconventions: end of line is a single character often known as '\n'.This differs from Windows, which ends lines with two charcters,'\n' and '\r'. Do not let your program get trapped by this!
PROGRAM NAME: gift1
INPUT FORMAT
| Line 1: | The single integer, NP | |||
| Lines 2..NP+1: | Each line contains the name of a group member | |||
| Lines NP+2..end: | NP groups of lines organized like this:
|
SAMPLE INPUT (file gift1.in)
5 dave laura owen vick amr dave 200 3 laura owen vick owen 500 1 dave amr 150 2 vick owen laura 0 2 amr vick vick 0 0
OUTPUT FORMAT
The output is NP lines, each with the name of a person followed bya single blank followed by the net gain or loss (final_money_value - initial_money_value) for that person. The names should be printed inthe same order they appear on line 2 of the input.
All gifts are integers. Each person gives the same integer amountof money to each friend to whom any money is given, and gives as muchas possible that meets this constraint. Any money not given is kept bythe giver.
SAMPLE OUTPUT (file gift1.out)
dave 302
laura 66
owen -359
vick 141
amr -150
解题思路：
该题考查的就是模拟能力，做起来其实 并没有难度，题目大意是说朋友之间相互送礼，当然容易忽略的一点是并不一定把所拥有的全部送人，为了表示公平，送出的
人中每个人得到的礼物数量是相同的，并且都是整数，这就要求我们判断每个人的礼物是否送完，结果输出是现在所拥有的礼物（即未送完的数量与得到礼物之和）与之前所拥
有的礼物的数量差。显而易见，用结构体来做，更为方便。
代码如下：
/*
ID:ayludon3
LANG: C++
TASK: gift1
*/
#include <iostream>
#include <fstream>
using namespace std;
struct gift
{
string name;
int cash;
int left;
int get;
int num;
};
int main()
{
ifstream fin ("gift1.in");
ofstream fout ("gift1.out");
gift person[12];
string str;
int m,n,i,j,k,l,p;
while(fin>>m)
{
k=m;
n=0;
for(i=0;i<k;i++)
{
fin>>person[i].name;
person[i].cash=person[i].get=person[i].left=person[i].num=0;
}
for(l=0;l<k;l++)
{
fin>>str;
for(j=0;j<k;j++)
if(person[j].name==str)
break;
fin>>person[j].cash>>person[j].num;
if(person[j].num!=0)
n=person[j].cash/person[j].num;
person[j].left=person[j].cash-person[j].num*n;
for(i=0;i<person[j].num;i++)
{
fin>>str;
for(p=0;p<k;p++)
if(person[p].name==str)
{
person[p].get+=n;
break;
}
}
}
for(i=0;i<k;i++)
fout<<person[i].name<<' '<<person[i].get+person[i].left-person[i].cash<<endl;
}
return 0;
}