S 1.1 gift C\ C++程序

题目：http://ace.delos.com/usacoprob2?a=w9IGt8u5aLX&S=gift1

 

题目：

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, NPLines 2..NP+1:Each line contains the name of a group memberLines NP+2..end:NP groups of lines organized like this:The first line in the group tells the person's name who will be giving gifts.The second line in the group contains two numbers: The initial 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, NG_i (0 ≤ NG_i ≤ NP-1).If NG_i is nonzero, each of the next NG_i lines lists the the name of a recipient of a gift.

SAMPLE INPUT (file gift1.in)

5davelauraowenvickamrdave200 3lauraowenvickowen500 1daveamr150 2vickowenlaura0 2amrvickvick0 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 302laura 66owen -359vick 141amr -150

题目翻译：

题目描述

对于一群（NP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 14 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。

[编辑]输入格式（gift1.in）

第 1 行: 人数NP,2<= NP<=10

第 2 行 到 第NP+1 行:这NP个在组里人的名字　一个名字一行

第NP＋2到最后：

这里的I段内容是这样组织的：

第一行是将会送出礼物人的名字。

第二行包含二个数字:　第一个是原有的钱的数目（在0到2000的范围里），第二个 NG_i 是将收到这个人礼物的人的个数 如果 NG_i 是非零的, 在下面 NG_i 行列出礼物的接受者的名字，一个名字一行。

[编辑]输出格式（gift1.out）

输出 NP 行

每行是一个的名字加上空格再加上收到的比送出的钱多的数目。

对于每一个人，他名字的打印顺序应和他在输入的2到NP＋1行中输入的顺序相同。所有的送礼的钱都是整数。

每个人把相同数目的钱给每位要接受礼物的朋友，而且尽可能多给，不能给出的钱由送礼者本人持有。

[编辑]重要提示

评测程序是基于UNIX的LINUX系统：换行符只是一个"\n",这与WINDOWS下换行符有两个字符" \n与\r"是不同的。
不要让你的程序踏入这个陷阱！

[编辑]样例输入

5davelauraowenvickamrdave200 3lauraowenvickowen500 1daveamr150 2vickowenlaura0 2amrvickvick0 0

[编辑]样例输出

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

 

 题目求解：

方法一：用结构体实现

/*ID:smilecl1LANG:CTASK:gift*//*TIME:2012年10月30日*/#include<stdio.h>#include<string.h>#define max 10#define max_length 15struct people{   char name[max_length];   int  money;};struct people people[max];int main(){ FILE *fin,*fout; char giver[max_length]={0}; char receiver[max_length]={0}; int np; int i=0; int j=0; int k=0; int giver_money=0; int giver_number=0; int money_temp=0; fin=fopen("input_file.in","r"); fout=fopen("output_file.out","w"); fscanf(fin, "%d",&np); for(i=0;i<max;i++)  {     // people[i].name={0};      people[i].money=0;  } for(i=0;i<np;++i) {     fscanf(fin,"%s",&people[i].name); }  for(i=0;i<np;++i)  {      k=0;      fscanf(fin,"%s%d%d",giver,&giver_money,&giver_number);      if(0==giver_money || 0==giver_number)        money_temp=0;       else        money_temp=giver_money/giver_number;       while(strcmp(people[k].name,giver))        k++;       people[k].money-=money_temp*giver_number;      for(j=0;j<giver_number;++j)      {          fscanf(fin,"%s",receiver);          k=0;          while(strcmp(people[k].name,receiver))            k++;          people[k].money+=money_temp;      }  }for(i=0;i<np;++i){    fprintf(fout,"%s %d\n",people[i].name,people[i].money);} fclose(fin); fclose(fout); exit(0);} 

 方法二：通过下标查找

/*ID:smilecl1LANG:CTASK:gift1*//*TIME:2012年10月30日*/#include <stdio.h>#include <stdlib.h>#include <string.h>int main(void){    FILE *in;    FILE *out;    int NP;    char name[10][14] = {0};    char giver[14] = {0};    char receiver[14] = {0};    int money[10]= {0};    int i=0;    int j=0;    int k=0;    int giver_number = 0;    int giver_money = 0;    int temp=0;    in=fopen("gift1.in", "r");    out=fopen("gift1.out", "w");    fscanf(in, "%d", &NP);    for(i=0; i!=NP; ++i)        fscanf(in,"%s",name[i]);    for(i=0; i<NP; ++i)    {        fscanf(in,"%s",giver);        fscanf(in,"%d%d",&giver_money,&giver_number);        if((0 == giver_money) || (0 == giver_number))            temp=0;        else            temp=giver_money/giver_number;        /*注意k要恢复0，不然第二次循环时，会出错。*/      k=0;       while(strcmp(name[k],giver))            k++;        money[k]=money[k]-temp*giver_number;        for(j=0; j < giver_number; j++)        {            fscanf(in,"%s", receiver);                       /* 注意k要恢复为0，不然k的值还是外循环的k的值，这个时候会导致receiver找不到数组定义的，因为              * 下标可能已经超过了receiver的下标，导致while死循环，k值最终会超出int的最大范围，导致出错，              * 编译却能通过。              */            k=0;              while(strcmp(name[k],receiver))            {                k++;            }            money[k]=money[k]+temp;        }    }    for(i=0; i != NP; i++)        fprintf(out,"%s %d\n", name[i], money[i]);    fclose(in);    fclose(out);    return 0;}
 
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