USACO题解——Section 1.2——Greedy Gift Givers
题目地址：https://train.usaco.org/usacoprob2?a=BGOMbIJsisd&S=gift1。
或者我的OJ网站，http://47.110.135.197/problem.php?id=5184。
题目
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 some or all of the other friends (although some might be cheap and give to no one). Likewise, each friend might or might not receive money from any or all of the other friends. Your goal is to deduce how much more money each person receives than they give.
The rules for gift-giving are potentially different than you might expect. Each person goes to the bank (or any other source of money) to get 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 7 among 2 friends would be 3 each for the friends with 1 left over – that 1 left over goes into the giver's "account". All the participants' gift accounts start at 0 and are decreased by money given and increased by money received.
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 money each person ends up with.
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 characters, '\n' and '\r'. Do not let your program get trapped by this!
PROGRAM NAME: gift1
INPUT FORMAT
| Line # | Contents | |||
|---|---|---|---|---|
| 1 | A single integer, NP | |||
| 2..NP+1 | Line i+1 contains the name of group member i | |||
| 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 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 starting 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
OUTPUT EXPLANATION
Five names: dave, laura, owen, vick, amr. Let's keep a table of the gives (money) each person 'has':
| dave | laura | owen | vick | amr |
| 0 | 0 | 0 | 0 | 0 |
| First, 'dave' splits 200 among 'laura', 'owen', and 'vick'. That comes to 66 each, with 2 left over | ||||
| -200+2 | +66 | +66 | +66 | 0 |
| → | ||||
| -198 | 66 | 66 | 66 | 0 |
| Second, 'owen' gives 500 to 'dave': | ||||
| -198+500 | 66 | 66-500 | 66 | 0 |
| → | ||||
| 302 | 66 | -434 | 66 | 0 |
| Third, 'amr' splits 150 between 'vick' and 'owen': | ||||
| 302 | 66 | -434+75 | 66+75 | -150 |
| → | ||||
| 302 | 66 | -359 | 141 | -150 |
| Fourth, 'laura' splits 0 between 'amr' and 'vick'; no changes: | ||||
| 302 | 66 | -359 | 141 | -150 |
| Finally, 'vick' gives 0 to no one: | ||||
| dave | laura | owen | vick | amr |
| 302 | 66 | -359 | 141 | -150 |
题目分析
题目含义
说了一堆，就是一堆人开始分钱，某个人拿出某个金额 X，分给 N 个指定名字的人，以此类推，直到游戏结束，然后输出最终所有人手上的钱。
算法归类
好吧一道标准的模拟题，没有什么难度。
数据范围
也就是说，最多 10 个人，每次拿出的钱是 2000。题目没有给出最多操作几次，也许这里是本题不够严谨的地方，所以解题的时候尽量开成 long long 的数据类型。
坑点
1、注意分法按照整数。什么意思，A 拿出 100 元分给 3 人，哪么每人分 33 元，也就是说 A 实际是拿出 99 元。
2、要注意最后输出的时候，要按照输入名字顺序输出。我开始第一次实现的时候使用了STL map这个数据结构，但是map是排序的。
3、文件结束如何表示。而不是样例的 0 0 表示结束，是 EOF。
AC参考代码
/*
ID: your_id_here
PROG: gift1
LANG: C++
*/
//Greedy Gift Givers
//https://train.usaco.org/usacoprob2?a=950pAjF03W0&S=gift1
#include
using namespace std;
struct PAIR {
string name;//名字
int deposit;//钱
};
const int MAXN = 10;
PAIR data[MAXN] = {};
int main() {
freopen("gift1.in", "r", stdin);
freopen("gift1.out", "w", stdout);
int np;
cin >> np;
for (int i=0; i> data[i].name;
data[i].deposit = 0;
}
string tmp;
int money, num;
while (cin >> tmp) {
cin >> money >> num;
if (0!=money && 0!=num) {
int quotient = money / num;//商
int remainder = money % num;//余数
//找到位置
for (int pos=0; pos> tmp;
for (int pos=0; pos> tmp;
}
}
}
//遍历
for (int i=0; i