1.1.2 USACO Task 'gift1': Greedy Gift Givers

https://train.usaco.org/usacoprob2?a=73eEaWy965w&S=gift1

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, '\r\ and '\n'. Do not let your program get trapped by this!

PROGRAM NAME: gift1

INPUT FORMAT

Line #

Contents

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:

The first line of each group tells the person's name who will be giving gifts.

The second line in the group contains two numbers:

The amount of money (in the range 0..2000) to be divided into gifts by the giver
NGi (0 ≤ NGi ≤ NP), the number of people to whom the giver will give gifts

If NGi is nonzero, each of the next NGi lines lists the name of a recipient of a gift; recipients are not repeated in a single giver's list.

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

题目描述

对于一群 nn 个要互送礼物的朋友，GY 要确定每个人送出的钱比收到的多多少。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 1414 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。

输入格式

第一行一个正整数 nn，表示人数。接下来nn 行，每行一个字符串表示人名。

接下来有 LL 段内容，对于每一段：第一行是将会送出礼物人的名字。
第二行包含二个非负整数，第一个是原有的钱的数目 ( \in [0,2000]∈[0,2000] )，第二个 g_igi 是将收到这个人礼物的人的个数如果 g_i \neq 0gi=0, 在下面 g_igi 行列出礼物的接受者的名字，一个名字一行。

输出格式

输出共 nn 行，每行输出一个人的名字和该人收到的钱比送出的钱多的数目。名字的顺序应该与输入第 22 行至 n+1n+1 行的顺序相同。

送出的钱永远是整数，即假设送礼人一次向 mm 人送出 nn 元，每个人应该得到 \lfloor n/m \rfloor⌊n/m⌋ 元。剩余未送出的钱应返还给送礼者。

输入输出样例

输入 #1复制

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

输出 #1复制

dave 302
laura 66
owen -359
vick 141
amr -150

说明/提示

【数据范围】
1\le n \le 101≤n≤10

解析：注意循环判断后需要break

/*
ID:  L
PROG: gift1
LANG: C++
*/
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
struct s1{string name;int money;
}s[12];
int main()
{freopen("gift1.in","r",stdin);freopen("gift1.out","w",stdout);ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);int n;int k = 0;cin >> n;for(int i = 1; i <= n; ++i)//输入姓名和钱 {cin >> s[i].name;s[i].money = 0;}//   for(int i = 1; i <= n; ++i)//    cout << s[i].name << endl;for(int i = 1;  i <= n; ++i){string name2,name3;cin >> name2;//送礼者 int a,b;cin >> a >> b;//将a元钱送给b个人 if(b == 0) continue;//需要特判，不然除数为0会报错 for(int u = 1; u <= n; ++u)//送礼者钱减少 {if(name2.compare(s[u].name) == 0){s[u].money = s[u].money - a + a%b;break;//  cout << s[u].name << " " << s[u].money;}}for(int j = 1; j <= b; ++j){cin >> name3;//接受礼物者 for(int u = 1; u <= n; ++u){if(name3.compare(s[u].name) == 0)//接受礼物者评分送礼者的钱 {s[u].money += a/b;break;//   cout << s[u].name << " " << s[u].money << endl;}}}}for(int u = 1; u <= n; ++u) {cout << s[u].name << " " << s[u].money << endl;}return 0;
}

使用map

map<string,int> mp; string 为key

mp["dave"] = 302

/*
ID: L
PROG: gift1
LANG: C++
*/
#include<iostream>
#include<cstring>
#include<string>
#include<map>
#include<cstdio>
using namespace std;
map<string,int> mp;
string s[12];
int s2[12];
int main()
{freopen("gift1.in","r",stdin);freopen("gift1.out","w",stdout);
//  ios::sync_with_stdio(false); // cin.tie(0);cout.tie(0);int n;int k = 0;cin >> n;for(int i = 1; i <= n; ++i)//输入姓名和钱 {cin >> s[i];s2[i] = 0;}// for(int i = 1; i <= n; ++i)//    cout << s[i].name << endl;for(int i = 1;  i <= n; ++i){string name2,name3;cin >> name2;//送礼者 int a,b;cin >> a >> b;//将a元钱送给b个人 if(b == 0) continue;//需要特判，不然除数为0会报错 mp[name2] = mp[name2] - a + a%b;for(int j = 1; j <= b; ++j){cin >> name3;//接受礼物者 mp[name3] += a/b;}}for(int i = 1; i <= n; ++i) {cout << s[i] << " " << mp[s[i]] << endl;}return 0;
}