USACO1.1.2 Greedy Gift Givers(贪婪的礼物送礼者)
2013-01-08 11:23
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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.
and '\r'. Do not let your program get trapped by this!
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.
题目大意:在这个问题中,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,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,你要确定每个人送出的礼物比收到的多多少。
解题思路:直接模拟即可,检索姓名时可以使用字符串哈希优化。
以下是我的代码:
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 printedin 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
题目大意:在这个问题中,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,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,你要确定每个人送出的礼物比收到的多多少。
解题思路:直接模拟即可,检索姓名时可以使用字符串哈希优化。
以下是我的代码:
/* USER:xingwen wang TASK:gift1 LANG:C */ #include<stdio.h> #include<string.h> int main() { freopen("gift1.in","r",stdin); freopen("gift1.out","w",stdout); int i,j,k,n,m,money,N,in[11]={0},out[11]={0}; char str[15],name[11][15]; scanf("%d",&n); for(i=0;i<n;i++) scanf("%s",name[i]); for(i=0;i<n;i++) { scanf("%s",&str); for(k=0;k<n;k++) if(strcmp(name[k],str)==0) break; scanf("%d%d",&money,&N); if(N!=0) { out[k]=money-money%N; m=N; while(m--) { scanf("%s",str); for(j=0;j<n;j++) if(strcmp(name[j],str)==0) { in[j]=in[j]+out[k]/N; break; } } } } for(i=0;i<n;i++) printf("%s %d\n",name[i],in[i]-out[i]); return 0; }
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