1.1.1---Greedy Gift Givers

Greedy Gift Givers

A group of NP (2 ≤ NP ≤ 10)uniquely named friends has decided to exchange gifts of money. Each of thesefriends might or might not give some money to any or all of the other friends.Likewise, each friend might or might not receive money from any or all of theother friends. Your goal in this problem is to deduce how much more money eachperson gives than they receive.

The rules for gift-giving are potentiallydifferent than you might expect. Each person sets aside a certain amount ofmoney to give and divides this money evenly among all those to whom he or sheis giving a gift. No fractional money is available, so dividing 3 among 2friends would be 1 each for the friends with 1 left over -- that 1 left overstays in the giver's "account".

In any group of friends, some people aremore giving than others (or at least may have more acquaintances) and somepeople have more money than others.

Given a group of friends, no one of whomhas a name longer than 14 characters, the money each person in the group spendson 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 they receive.

IMPORTANT NOTE

The grader machine is a Linux machine thatuses standard Unix conventions: end of line is a single character often knownas '\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: Thefirst line in the group tells the person's name who will be giving gifts. 

The second line in the group contains twonumbers: The initial amount of money (in the range 0..2000) to be divided upinto gifts by the giver and then the number of people to whom the giver willgive gifts, NGi (0 ≤ NGi ≤ NP-1). 

If NGi is nonzero, each of the next NGilines lists the the name of a recipient of a gift. 

 

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 nameof a person followed by a single blank followed by the net gain or loss (final_money_value- initial_money_value) for that person. The names should be printed in the sameorder they appear on line 2 of the input.

All gifts are integers. Each person givesthe same integer amount of money to each friend to whom any money is given, andgives as much as possible that meets this constraint. Any money not given iskept by the giver.

SAMPLE OUTPUT (file gift1.out)

dave 302

laura 66

owen -359

vick 141

amr -150

分析：

贪婪的礼物送礼者

译 by tim green

对于一群要互送礼物的朋友(2 ≤ NP ≤ 10)，你要确定每个人送出的礼物比收到的多多少(and vice versa for those who view gift giving with cynicism)。

在这一个问题中，每个人都准备了一些钱来送礼物，而这些钱将会被平均分给那些将收到他的礼物的人。

然而，在任何一群朋友中，有些人将送出较多的礼物(可能是因为有较多的朋友)，有些人有准备了较多的钱。

给出一群朋友，没有人的名字会长于 14 字符，给出每个人将花在送礼上的钱，和将收到他的礼物的人的列表，

请确定每个人收到的比送出的钱多的数目。

IMPORTANT NOTE

测试系统是 Linux 符合标准的 Unix 的协定。

用'\n'作为行的结束。

这和 Windows 系统用'\n' 和 '\r'作为行的结束是不同的。

你的程序不要被这困住了。

PROGRAM NAME: gift1

INPUT FORMAT 第 1 行: 人数NP,2<=NP<=10

第 2到 NP+1 行:

这NP个在组里人的名字一个名字一行

第NP＋2到最后：

这里的NP段内容是这样组织的：

第一行是将会送出礼物人的名字。

第二行包含二个数字: 第一个是原有的钱的数目（在0到2000的范围里），第二个NGi是将收到这个送礼者礼物的人的个数如果 NGi 是非零的, 在下面 NGi 行列出礼物的接受者的名字，一个名字一行。

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

输出 NP 行

每行是一个的名字加上空格再加上收到的比送出的钱多的数目。

对于每一个人，他名字的打印顺序应和他在输入的2到NP＋1行中输入的顺序相同。所有的送礼的钱都是整数。

每个人把相同数目的钱给每位要送礼的朋友，而且尽可能多给，不能给出的钱被送礼者自己保留。

SAMPLE OUTPUT (file gift1.out)

dave 302

laura 66

owen -359

vick 141

amr -150

/*

ID: ******

PROG: gift1

LANG: C++

*/

#include <iostream>

#include <fstream>

#include <string>

#include <map>

using namespace std;

string st[12];

int main() {

   ofstream fout ("gift1.out");

   ifstream fin ("gift1.in");

         map<string,int>m;

         intn,i,temp1,temp2,j;

         stringtemp;

   string a, b;

         fin>>n;

         for(i=0;i<n;i++)

                   {

                            fin>>st[i];//读入朋友圈

                            m[st[i]]=0;

         }

        

         for(i=0;i<n;i++)

         {

         fin>>temp;//读入送礼人名字

         fin>>temp1>>temp2;//读入起始资金和送礼对象数量

         if(!temp2)

                   m[temp]=m[temp];//排除不送人礼物的吝啬鬼情况

         else

         m[temp]-=(temp1-(temp1%temp2));//在结果里减去送出去的钱

         j=temp2;

         while(temp2--)

         {

                   fin>>temp;

                   m[temp]+=(temp1/j);//在结果中加上得到的钱

 

         }

         }

         for(i=0;i<n;i++)

                   fout<<st[i]<<""<<m[st[i]]<<endl;//写文件

   return 0;

}