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:
|
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
===================================================
比如样例，有5个人，分别是dave，laura，owen，vick，amr。
接下来每输入一次操作，都是先输入被分享的那个人，再输入其被瓜分的money数目，以及瓜分他钱的人数num，接下来的num行即是瓜分其钱财的人名
考虑几种非常规情况，特别是money不为0，num为0的时候，此时被瓜分的人不能得到钱，而是全数扣除
常规情况下就是被瓜分的人会失去money总数并增加被瓜分后剩余的(money % num)。
AC代码:
1 /*
2 ID: luopengting
3 PROG: gift1
4 LANG: C++
5 */
6 #include <iostream>
7 #include <cstdio>
8 #include <vector>
9 using namespace std;
10 const int maxn = 2005;
11 int score[maxn];
12 int main()
13 {
14
15 freopen("gift1.in", "r", stdin);
16 freopen("gift1.out", "w", stdout);
17
18 vector<string> person;
19 int num, t = 0, n, money;
20 string s;
21 //memset(sum, 0, sizeof(sum));
22 cin >> n;
23 for(int i = 0; i < n; i++)
24 {
25 cin >> s;
26 person.push_back(s);
27 score[i] = 0;
28 }
29
30 for(int k = 0; k < n; k++)
31 {
32 cin >> s;
33 cin >> money >> num;
34 if(money == 0 && num != 0)
35 {
36 for(int i = 0; i < num; i++)
37 {
38 cin >> s;
39 }
40 continue;
41 }
42 for(int i = 0; i < n; i++)
43 {
44 if(person[i] == s)
45 {
46 if(num)
47 {
48 score[i] += (money % num) - money;
49 }
50 else //当money不为0，num为0的时候！！！考虑这个就过了！！
51 {
52 score[i] -= money;
53 }
54 }
55 }
56 for(int i = 0; i < num; i++)
57 {
58 cin >> s;
59 for(int i = 0; i < n; i++)
60 {
61 if(person[i] == s)
62 {
63 score[i] += money / num;
64 }
65 }
66 }
67 }
68 for(int i = 0; i < n; i++)
69 {
70 cout << person[i] << " " << score[i] << endl;
71 }
72 return 0;
73 }
为了玩map，也写了下面的代码，不过这份代码不能完成题目当中的输出，因为其顺序是字典序排列 纯属娱乐
1 #include <iostream>
2 #include <cstdio>
3 #include <vector>
4 #include <map>
5 using namespace std;
6 int main()
7 {
8 #ifdef LUOPENGTING_JUDGE
9 freopen("gift1.in", "r", stdin);
10 freopen("gift1,out", "w", stdout);
11 #endif // LUOPENGTING_JUDGE
12 map<string, int> person;
13 map<string, int>::iterator iter;
14 int num, t = 0, n, money;
15 string s;
16 //memset(sum, 0, sizeof(sum));
17 cin >> n;
18 for(int i = 0; i < n; i++)
19 {
20 cin >> s;
21 person.insert(pair<string, int>(s, t));
22 }
23
24 while(cin >> s)
25 {
26 cin >> money >> num;
27 if(money == 0 && num != 0)
28 {
29 for(int i = 0; i < num; i++)
30 {
31 cin >> s;
32 }
33 continue;
34 }
35 if(money == 0 && num == 0)
36 {
37 break;
38 }
39
40 iter = person.lower_bound(s);
41 iter -> second = iter ->second - money + (money % num);
42 for(int i = 0; i < num; i++)
43 {
44 cin >> s;
45 iter = person.lower_bound(s);
46 iter -> second += (money / num);
47 }
48 }
49 map<string, int>::reverse_iterator it;
50 for(it = person.rbegin(); it != person.rend(); it++)
51 {
52 cout << it->first << " " << it->second<<endl;
53 }
54 return 0;
55 }
View Code
原文链接: https://www.cnblogs.com/imLPT/p/4132806.html
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