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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.
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!
| 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:
|
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
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.
dave 302 laura 66 owen -359 vick 141 amr -150
这道题在codeforces上面做到过思路一样的 大概就是map的应用吧
就是有很多人互相送礼物 每个人送礼物的钱是平均分给其他人的 最后通过每个人送出去的钱和收到的礼物的价钱的比较 求出最终每个人的收获 如果不能平均分给其他人 取模的值留给自己
其实就是map<string, int> string存人名 int存送出和收获的礼物的价钱 最后输出结果就好了 感觉会用map这道题可以很快做出来 不会用的话就结构体也不会用很久
这道题收获最深有以下两点
1. USACO真的是很多题目的来源参考
2. STL真是个好东西
/*
ID: Alpha Augur
PROG: gift1
LANG: C++
*/
#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
string str[40];
int main(){
freopen("gift1.in", "r", stdin);
freopen("gift1.out", "w", stdout);
int n;
map<string, int> na;
cin >> n;
for(int i = 0; i < n; ++i){
cin >> str[i];
}
for(int i = 0; i < n; ++i){
string s;
int num, m;
cin >> s;
cin >> num >> m;
na[s] -= num;
if(m != 0){
na[s] += num % m;
}
for(int j = 0; j < m; ++j){
cin >> s;
na[s] += num / m;
}
}
for(int i = 0; i < n; ++i){
cout << str[i] << " " << na[str[i]] << endl;
}
return 0;
}
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【USACO1.1.2】Greedy Gift Givers（map）
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原文地址：http://blog.csdn.net/u012431590/article/details/46757249
