Posted on 2008-04-08 19:12
Codeboy 阅读(561)
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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.
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:
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Each line contains the name of a group member
|
|
Lines NP+2..end:
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NP groups of lines organized like this:
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The first line in the group tells the person’s name who will be giving gifts.
|
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The second line in the group contains two numbers: The initail 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).
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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
我的解答：
题目意思比较简单，就是有NP个人，每个人都向其他人送钱，同时他又会收到别人送给他的钱。题目要求计算每个人收到的钱比送出的钱多多少。
算法比较明了，但要注意一些细节问题，在送钱的时候，会给出一个总数和要分的分数，分的时候只保留整数，不会出现小数。
#include <fstream>
using namespace std;
struct NP
{
char name[15];
int give;
int receive;
};
inline int find(NP member[], int n, char name[])
{
for (int i = 0; i < n; i++)
if (!strcmp(member[i].name, name)) return i;
}
int main(void)
{
ofstream fout("gift1.out");
ifstream fin("gift1.in");
NP member[10];
int np;
fin >> np;
for (int i = 0; i < np; i++) {
fin >> member[i].name;
member[i].give = member[i].receive= 0;
}
for (int i = 0; i < np; i++) {
char name[15];
int money, num;
fin >> name;
fin >> money >> num;
if (num) {
member[find(member, np, name)].give = money / num * num;
for (int j = 0; j < num; j++) {
fin >> name;
member[find(member, np, name)].receive += money / num;
}
}
}
for (int i = 0; i < np; i++)
fout << member[i].name << ' ' << (member[i].receive - member[i].give) << endl;
return 0;
}官方解答：
Analysis:Greedy Gift Givers
The hardest part about this problem is dealing with the strings representing people's names.
We keep an array of Person structures that contain their name and how much money they give/get.
The heart of the program is the lookup() function that, given a person's name, returns their Person structure. We add new people with addperson().
Note that we assume names are reasonably short.
#include <stdio.h>
#include <string.h>
#include <assert.h>
#define MAXPEOPLE 10
#define NAMELEN 32
typedef struct Person Person;
struct Person {
char name[NAMELEN];
int total;
};
Person people[MAXPEOPLE];
int npeople;
void addperson(char *name)
{
assert(npeople < MAXPEOPLE);
strcpy(people[npeople].name, name);
npeople++;
}
Person* lookup(char *name)
{
int i;
/**//* look for name in people table */
for(i=0; i<npeople; i++)
if(strcmp(name, people[i].name) == 0)
return &people[i];
assert(0); /**//* should have found name */
}
int main(void)
{
char name[NAMELEN];
FILE *fin, *fout;
int i, j, np, amt, ng;
Person *giver, *receiver;
fin = fopen("gift1.in", "r");
fout = fopen("gift1.out", "w");
fscanf(fin, "%d", &np);
assert(np <= MAXPEOPLE);
for(i=0; i<np; i++) {
fscanf(fin, "%s", name);
addperson(name);
}
/**//* process gift lines */
for(i=0; i<np; i++) {
fscanf(fin, "%s %d %d", name, &amt, &ng);
giver = lookup(name);
for(j=0; j<ng; j++) {
fscanf(fin, "%s", name);
receiver = lookup(name);
giver->total -= amt/ng;
receiver->total += amt/ng;
}
}
/**//* print gift totals */
for(i=0; i<np; i++)
printf("%s %d\n", people[i].name, people[i].total);
exit (0);
}