USACO 1.1 Greedy Gift Givers
题目来源：USACO 1.1
原题目：
Greedy Gift Givers
A group of NP (2 ≤ NP ≤ 10) uniquely named friends has decided to exchange gifts of money.
Each of these friends 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 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 you might expect. Each person sets
aside a certain amount of money to give and divides this money evenly among all those to
whom he or she is giving a gift. No fractional money is available, so dividing 3 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 more money than others.
Given a group of friends, no one of whom has a name longer than 14 characters, the money
each person in the group spends on gifts, and a (sub)list of friends to whom each person
gives gifts, determine how much more (or less) each person in the group gives than they
receive.
IMPORTANT NOTE
The grader machine is a Linux machine that uses standard Unix conventions: 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: The first line in the group tells
the person's name who will be giving gifts.
The second line in the group contains two numbers: The initial amount of money (in the
range 0..2000) to be divided up into gifts by the giver and then the number of people to
whom the giver will give gifts, NGi (0 ≤ NGi ≤ NP-1).
If NGi is nonzero, each of the next NGi lines 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 name of 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 same order they appear on line 2 of the input.
All gifts are integers. Each person gives the same integer amount of money to each friend
to whom any money is given, and gives as much as possible that meets this constraint. Any
money not given is kept by the giver.
SAMPLE OUTPUT (file gift1.out)
dave 302
laura 66
owen -359
vick 141
amr -150
译题：
对于一群要互送礼物的朋友，你要确定每个人收到的礼物比送出的多多少,反之对别人来看礼物的人。
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。
格式：
程序名称: gift1
(file gift1.in)
第 1 行: 人数NP,2<= NP<=10
第 2到 J+1 行:这NP个在组里人的名字 一个名字一行
第J＋2到最后：
这里的I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。
输出格式:
(file gift1.out)
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，不能给出的钱送礼者自己保留(算进送
礼者本人收入部分)。
View Code
1 /*
2 ID:flyings3
3 PROG:gift1
4 LANG:C++
5 */
6
7 #include <fstream>
8 #include <cstring> //用到C++字符串了所以必须添加这个头文件 比赛可用
9
10 using namespace std;
11
12 short n;
13
14 int main()
15 {
16 ifstream fin("gift1.in");
17 fin>>n;
18 string name[10];
19 short fn[10];
20 short given[10];
21 short gotten[10]={0};//这个必须初始化
22 short i;
23 for (i=0; i<n; fin>>name[i++]);
24 short j,k;
25 string t;
26 short x;
27 for (i=0; i<n; i++)
28 {
29 fin>>t;
30 for (x=0; x<n; x++)
31 if (t==name[x])
32 break;
33 //根据姓名搜索匹配 x即为送礼者的编号
34 fin>>given[x];
35 fin>>fn[x];
36 for (j=0; j<fn[x]; j++)
37 {
38 fin>>t;
39 for (k=0; k<n; k++)
40 if (t==name[k])
41 break;
42 //根据姓名搜索匹配 k为收礼者编号
43 gotten[k]+=(given[x]/fn[x]);
44 }
45 }
46 fin.close();
47
48 ofstream fout("gift1.out");
49 for (i=0; i<n; i++)
50 {
51 if (fn[i]==0) //防止被0除
52 fout<<name[i]<<" "<<(gotten[i]+given[i])<<endl;
53 else
54 fout<<name[i]<<" "<<(given[i]%fn[i]+gotten[i]-given[i])<<endl;
55 //注意送礼余下的钱算进收益
56 }
57 fout.close();
58 return 0;
59 }
