Problem Statement

    

In most states, gamblers can choose from a wide variety of different lottery games. The rules of a lottery are defined by two integers (choices and blanks) and two boolean variables (sorted and unique). choices represents the highest valid number that you may use on your lottery ticket. (All integers between 1 and choices, inclusive, are valid and can appear on your ticket.) blanks represents the number of spots on your ticket where numbers can be written.

The sorted and unique variables indicate restrictions on the tickets you can create. If sorted is set to true, then the numbers on your ticket must be written in non-descending order. If sorted is set to false, then the numbers may be written in any order. Likewise, if unique is set to true, then each number you write on your ticket must be distinct. If unique is set to false, then repeats are allowed.

Here are some example lottery tickets, where choices = 15 and blanks = 4:

  • {3, 7, 12, 14} -- this ticket is unconditionally valid.
  • {13, 4, 1, 9} -- because the numbers are not in nondescending order, this ticket is valid only if sorted = false.
  • {8, 8, 8, 15} -- because there are repeated numbers, this ticket is valid only if unique = false.
  • {11, 6, 2, 6} -- this ticket is valid only if sorted = false and unique = false.

Given a list of lotteries and their corresponding rules, return a list of lottery names sorted by how easy they are to win. The probability that you will win a lottery is equal to (1 / (number of valid lottery tickets for that game)). The easiest lottery to win should appear at the front of the list. Ties should be broken alphabetically (see example 1).

Definition

    
Class: Lottery
Method: sortByOdds
Parameters: vector <string>
Returns: vector <string>
Method signature: vector <string> sortByOdds(vector <string> rules)
(be sure your method is public)
    
 

Constraints

- rules will contain between 0 and 50 elements, inclusive.
- Each element of rules will contain between 11 and 50 characters, inclusive.
- Each element of rules will be in the format "<NAME>:_<CHOICES>_<BLANKS>_<SORTED>_<UNIQUE>" (quotes for clarity). The underscore character represents exactly one space. The string will have no leading or trailing spaces.
- <NAME> will contain between 1 and 40 characters, inclusive, and will consist of only uppercase letters ('A'-'Z') and spaces (' '), with no leading or trailing spaces.
- <CHOICES> will be an integer between 10 and 100, inclusive, with no leading zeroes.
- <BLANKS> will be an integer between 1 and 8, inclusive, with no leading zeroes.
- <SORTED> will be either 'T' (true) or 'F' (false).
- <UNIQUE> will be either 'T' (true) or 'F' (false).
- No two elements in rules will have the same name.

Examples

0)  
    
{"PICK ANY TWO: 10 2 F F"
,"PICK TWO IN ORDER: 10 2 T F"
,"PICK TWO DIFFERENT: 10 2 F T"
,"PICK TWO LIMITED: 10 2 T T"}
Returns:
{ "PICK TWO LIMITED",
"PICK TWO IN ORDER",
"PICK TWO DIFFERENT",
"PICK ANY TWO" }

The "PICK ANY TWO" game lets either blank be a number from 1 to 10. Therefore, there are 10 * 10 = 100 possible tickets, and your odds of winning are 1/100.

The "PICK TWO IN ORDER" game means that the first number cannot be greater than the second number. This eliminates 45 possible tickets, leaving us with 55 valid ones. The odds of winning are 1/55.

The "PICK TWO DIFFERENT" game only disallows tickets where the first and second numbers are the same. There are 10 such tickets, leaving the odds of winning at 1/90.

Finally, the "PICK TWO LIMITED" game disallows an additional 10 tickets from the 45 disallowed in "PICK TWO IN ORDER". The odds of winning this game are 1/45.

1)  
    
{"INDIGO: 93 8 T F",
"ORANGE: 29 8 F T",
"VIOLET: 76 6 F F",
"BLUE: 100 8 T T",
"RED: 99 8 T T",
"GREEN: 78 6 F T",
"YELLOW: 75 6 F F"}
Returns: { "RED",  "ORANGE",  "YELLOW",  "GREEN",  "BLUE",  "INDIGO",  "VIOLET" }

Note that INDIGO and BLUE both have the exact same odds (1/186087894300). BLUE is listed first because it comes before INDIGO alphabetically.

2)  
    
{}
Returns: { }

Empty case

总的来说此题不难,就是组合数学题

#include <vector>
#include <string>
#include <stdlib.h>
#include <stdint.h>
#include <stdio.h> using namespace std; class Lottery {
public:
vector <string> sortByOdds(vector <string>);
void compileRule(string, string&, uint64_t&, uint64_t&, int&, int&);
uint64_t computeOdd(uint64_t, uint64_t, int, int);
vector<string> ascendOrder(uint64_t* ,string[], int);
}; vector <string> Lottery::sortByOdds(vector <string> rules) {
int len=rules.size();
int i;
string names[len];
uint64_t choices[len];
uint64_t blanks[len];
uint64_t odds[len];
int sorted[len];
int unique[len];
vector<string> result;
if(len==0){
return result;
}
for(i=0;i<len;i++){
compileRule(rules.at(i), names[i], choices[i], blanks[i], sorted[i], unique[i]);
}
for(i=0;i<len;i++){
odds[i] = computeOdd(choices[i], blanks[i], sorted[i], unique[i]);
}
result=ascendOrder(odds, names, len);
return result;
} void Lottery::compileRule(string rule, string& name, uint64_t& choice, uint64_t& blank, int& sorted, int&unique){
char *a = (char*)malloc(sizeof(char));
choice = 0;
blank = 0;
//extract name until find an :
string::iterator iter;
for(iter=rule.begin();*iter!=':';iter++){
*a=*iter;
name.append((const char*)a);
}
iter=iter+2;
//extract choice
for(;*iter!=' ';iter++){
choice=choice*10+*iter-48;
}
iter++;
//extract blank
for(;*iter!=' ';iter++){
blank=blank*10+*iter-48;
}
//printf("blank %d\n",blank);
iter++;
if(*iter=='T') sorted=1;
else sorted=0;
iter=iter+2;
if(*iter=='T') unique=1;
else unique=0;
//printf("choice %d, blank %d, sorted %d, unique %d\n",choice,blank, sorted, unique);
}
uint64_t Lottery::computeOdd(uint64_t choice, uint64_t blank, int sorted, int unique){
int i;
uint64_t result=1;
uint64_t tresult=1;
if(sorted == 0 && unique == 0){
for(i = 0;i<blank;i++){
result=result*choice;
}
}
if(sorted==1 && unique==0){
for(i = 0;i<blank;i++){
result=result*choice;
}
for(i=0;i<blank;i++){
tresult=tresult*(choice-i);
}
for(i=0;i<blank;i++){
tresult=tresult/(i+1);
}
result=result-tresult;
}
if(sorted==0 && unique==1){
for(i=0;i<blank;i++){
result=result*(choice-i);
}
}
if(sorted==1 && unique==1){
for(i=0;i<blank;i++){
result=result*(choice-i);
}
for(i=0;i<blank;i++){
result=result/(i+1);
}
}
printf("odd %d\n",result);
//printf("choice %d, blank %d, sorted %d, unique %d, odd %d\n",choice,blank, sorted, unique,result);
return result;
} vector<string> Lottery::ascendOrder(uint64_t* odd, string* names, int len){
uint64_t temp;
int order[len];
int i,j, torder;
vector<string> result;
for(i=0;i<len;i++)
order[i]=i;
for(i=0;i<len-1;i++){
for(j=0;j<len-1-i;j++){
if (odd[j]>odd[j+1]){
temp=odd[j];
odd[j]=odd[j+1];
odd[j+1]=temp;
torder=order[j];
order[j]=order[j+1];
order[j+1]=torder;
}
}
}
for(i=0;i<len;i++)
printf("%d ",odd[i]);
for(i=0;i<len;i++)
printf("%d ",order[i]);
for(i=0;i<len;i++){
result.push_back(names[order[i]]);
}
return result;
}

  

topcoder算法练习2的更多相关文章

  1. topcoder算法练习3

    SRM144 DIV1 1100 point Problem Statement      NOTE: There are images in the examples section of this ...

  2. ITWorld:2014年全球最杰出的14位编程天才

    近日,ITWorld 整理全球最杰出的 14 位程序员,一起来看下让我们膜拜的这些大神都有哪些?(排名不分先后) 1.Jon Skeet 个人名望:程序技术问答网站 Stack Overflow 总排 ...

  3. BFS/DFS算法介绍与实现(转)

    广度优先搜索(Breadth-First-Search)和深度优先搜索(Deep-First-Search)是搜索策略中最经常用到的两种方法,特别常用于图的搜索.其中有很多的算法都用到了这两种思想,比 ...

  4. IT求职中,笔试、面试的算法准备

    PS:此文章为转载,源地址:http://www.newsmth.net/nForum/#!article/CoderInterview/849     作者应该是在美国进行的笔试面试,感觉面试的的公 ...

  5. *[topcoder]LCMSetEasy

    http://community.topcoder.com/stat?c=problem_statement&pm=13040 DFS集合全排列+LCM和GCD.但事实上,有更简单的算法,列在 ...

  6. *[topcoder]LittleElephantAndBalls

    http://community.topcoder.com/stat?c=problem_statement&pm=12758&rd=15704 topcoder的题经常需要找规律,而 ...

  7. [topcoder]KingdomReorganization

    http://community.topcoder.com/stat?c=problem_statement&pm=11282&rd=14724 这道题是最小生成树,但怎么转化是关键. ...

  8. [topcoder]ActivateGame

    http://community.topcoder.com/stat?c=problem_statement&pm=10750&rd=14153 http://apps.topcode ...

  9. [topcoder]BestRoads

    http://community.topcoder.com/stat?c=problem_statement&pm=10172&rd=13515 http://community.to ...

随机推荐

  1. (转载)如何优化MySQL insert性能

    (转载)http://blog.csdn.net/tigernorth/article/details/8094277 对于一些数据量较大的系统,面临的问题除了是查询效率低下,还有一个很重要的问题就是 ...

  2. Windows玩转Docker(二):运行whalesay image

    docker官网site:http://www.docker.com/ 参照site:https://docs.docker.com/windows/step_three/ docker安装参照: h ...

  3. Java GC 专家系列5:Java应用性能优化的原则

    本文是GC专家系列中的第五篇.在第一篇理解Java垃圾回收中我们学习了几种不同的GC算法的处理过程,GC的工作方式,新生代与老年代的区别.所以,你应该已经了解了JDK 7中的5种GC类型,以及每种GC ...

  4. HBase 简介(强烈推荐看)

    本博文的主要内容有: .HBase定义 .HBase 的特点 .HBase 访问接口  .HBase 存储结构 .HBase设计 .HBase安装 .HBase shell操作  .输入 help 可 ...

  5. PHP——四种基本排序算法

    分别用冒泡排序法,快速排序法,选择排序法,插入排序法将下面数组中的值按照从小到大的顺序进行排序. $arr(1,43,54,62,21,66,32,78,36,76,39); 1. 冒泡排序 思路分析 ...

  6. Docker的基本操作

    容器基本操作 1.启动容器 $docker run image [COMMAND] [ARG…] run在新容器中执行命令 2.启动交互式容器 $docker run -i -t IMAGE /bin ...

  7. iOS开发总结-UIWebView 集成 浏览器

    // // detailWebViewController.m // BJ // // Created by shirenfeng on 16/11/6. // Copyright © 2016年 c ...

  8. 坚持c++,真正掌握c++(2)

    在c++中对c中的输入输出进行了扩展,採用了面向对象的设计方法设计了c++中的输入输出(IO).输入输出依照操作的对象分类可分为:1. 标准IO(对计算机的键盘或者显示器进行读写操作).2. 文件IO ...

  9. TopCoder SRMS 1 字符串处理问题 Java题解

    Problem Statement   Let's say you have a binary string such as the following: 011100011 One way to e ...

  10. linux 内核分析+使用SystemTap调试新增内核模块

    http://blog.chinaunix.net/uid/14528823/list/1.html?cid=189394