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. [IoLanguage]Io Programming Guide[转]

    Io Programming Guide     Introduction Perspective Getting Started Downloading Installing Binaries Ru ...

  2. vijosP1014 旅行商简化版

    vijosP1014 旅行商简化版 链接:https://vijos.org/p/1014 [思路] 双线DP. 设ab,ab同时走.用d[i][j]表示ab所处结点i.j,且定义i>j,则有转 ...

  3. lightoj 1021 - Painful Bases 状态压缩

    题目链接:http://lightoj.com/volume_showproblem.php?problem=1021 #include<cstring> #include<cstd ...

  4. 从spark架构中透视job

    本博文的主要内容如下: 1.通过案例观察Spark架构 2.手动绘制Spark内部架构 3.Spark Job的逻辑视图解析 4.Spark Job的物理视图解析 1.通过案例观察Spark架构 sp ...

  5. 第十七章、程序管理与 SELinux 初探

    ---恢复内容开始--- 什么是程序 (process) 在 Linux 底下所有的命令与你能够进行的动作都与权限有关, 而系统依据UID/GID以及文件的属性相关性判定你的权限!在 Linux 系统 ...

  6. Skype的故事:几乎所有风投都想投 犯罪分子洗钱必备

    Skype的故事:几乎所有风投都想投 犯罪分子洗钱必备 转载自: http://news.chinaventure.com.cn/11/7/1381032922.shtml 今年是 Skype 网络电 ...

  7. sql宽字节注入,绕过单引号

    参加下面: http://leapar.lofter.com/post/122a03_3028a9 http://huaidan.org/archives/2268.html https://ilia ...

  8. c#基础语言编程-按值类型和引用类型传递参数

    引言 在介绍传递参数介绍前,请明白参数类型和传递方式是不同的.传递方式分为按值类型和引用类型传递参数.参数类型有值类型和引用类型,这个和C++是不同的.这里的传递方式对应c++中的深复制和浅复制. 两 ...

  9. UINavigationBar-使用总结

    多视图应用程序中,我们常常使用到自定义UINavigationBar来完成导航条的设置.   1.获取导航条   UINavigationBar *navBar = self.navigationCo ...

  10. Notepad++中的复活节彩蛋(easter egg)

    Notepad++是windows下的一个强大的文本编辑器; 它免费,开源, 比windows自带的记事本强大百倍; 我自己的感觉: 1.打开超大的文本文件,比方100M大小的文本格式的数据,记事本会 ...