2013
12-26

# The SetStack Computer

Background from Wikipedia: �et theory is a branch of mathematics created principally by the German mathematician Georg Cantor at the end of the 19th century. Initially controversial, set theory has come to play the role of a foundational theory in modern mathematics, in the sense of a theory invoked to justify assumptions made inmathematics concerning the existence of mathematical objects (such as numbers or functions) and their properties. Formal versions of set theory also have a foundational role to play as specifying a theoretical ideal of mathematical rigor in proofs.?Given this importance of sets, being the basis of mathematics, a set of eccentric theorist set off to construct a supercomputer operating on sets instead of numbers. The initial Set-Stack Alpha is under construction, and they need you to simulate it in order to verify the operation of the prototype.

The computer operates on a single stack of sets, which is initially empty. After each operation, the cardinality of the topmost set on the stack is output. The cardinality of a set S is denoted |S| and is the number of elements in S. The instruction set of the SetStack Alpha is PUSH, DUP, UNION, INTERSECT, and ADD.
?PUSH will push the empty set {} on the stack.
?DUP will duplicate the topmost set (pop the stack, and then push that set on the stack twice).
?UNION will pop the stack twice and then push the union of the two sets on the stack.
?INTERSECT will pop the stack twice and then push the intersection of the two sets on the stack.
?ADD will pop the stack twice, add the first set to the second one, and then push the resulting set on the stack.
For illustration purposes, assume that the topmost element of the stack is A = {{}, {{}}} and that the next one is B = {{}, {{{}}}}.
For these sets, we have |A| = 2 and |B| = 2. Then:
?UNION would result in the set { {}, {{}}, {{{}}} }. The output is 3.
?INTERSECT would result in the set { {} }. The output is 1.
?ADD would result in the set { {}, {{{}}}, {{},{{}}} }. The output is 3.

An integer 0 <= T <= 5 on the first line gives the cardinality of the set of test cases. The first line of each test case contains the number of operations 0 <= N <= 2 000. Then follow N lines each containing one of the five commands. It is guaranteed that the SetStack computer can execute all the commands in the sequence without ever popping an empty stack.

An integer 0 <= T <= 5 on the first line gives the cardinality of the set of test cases. The first line of each test case contains the number of operations 0 <= N <= 2 000. Then follow N lines each containing one of the five commands. It is guaranteed that the SetStack computer can execute all the commands in the sequence without ever popping an empty stack.

2
9
PUSH
DUP
PUSH
DUP
DUP
UNION
5
PUSH
PUSH
PUSH
INTERSECT

0
0
1
0
1
1
2
2
2
***
0
0
1
0
0
***

1 题目给定5种操作，每次输出栈顶集合的元素的个数

#include<set>
#include<map>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int N = 20;
const int MAXN = 2010;

int cnt;
stack<set<int> >stk;
map<set<int> , int>mp;
set<int>s1 , s2;

void pop(){
s1 = stk.top();
stk.pop();

s2 = stk.top();
stk.pop();
}
// push
void Push(){
set<int>s;
stk.push(s);
puts("0");
}
// dup
void Dup(){
set<int>s;
s = stk.top();
stk.push(s);
printf("%d\n" , s.size());
}
// union
void Union(){
pop();
//
set<int>::iterator it;
for(it = s1.begin() ; it != s1.end() ; it++)
s2.insert(*it);
stk.push(s2);
printf("%d\n" , s2.size());
}
// Intersect
void Intersect(){
pop();
//
set<int>s3;
set<int>::iterator it;
for(it = s1.begin() ; it != s1.end() ; it++){
if(s2.find(*it) != s2.end())
s3.insert(*it);
}
stk.push(s3);
printf("%d\n" , s3.size());
}
pop();
//
if(s1.empty())
s2.insert(0);
else{
if(!mp[s1])
mp[s1] = cnt++;
s2.insert(cnt++);
}
stk.push(s2);
printf("%d\n" , s2.size());
}

int main(){
int Case , n;
char str[N];
scanf("%d" , &Case);
while(Case--){
scanf("%d" , &n);
while(!stk.empty())
stk.pop();
cnt = MAXN;
mp.clear();
while(n--){
scanf("%s" , str);
if(str[0] == 'P')
Push();
else if(str[0] == 'D')
Dup();
else if(str[0] == 'U')
Union();
else if(str[0] == 'I')
Intersect();
else
}
puts("***");
}
return 0;
}

1. 题本身没错，但是HDOJ放题目的时候，前面有个题目解释了什么是XXX定律。
这里直接放了这个题目，肯定没几个人明白是干啥

2. #!/usr/bin/env python
def cou(n):
arr =
i = 1
while(i<n):
arr.append(arr[i-1]+selfcount(i))
i+=1
return arr[n-1]

def selfcount(n):
count = 0
while(n):
if n%10 == 1:
count += 1
n /= 10
return count

3. 这道题目虽然简单，但是小编做的很到位，应该会给很多人启发吧！对于面试当中不给开辟额外空间的问题不是绝对的，实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟，今天看到小编这篇文章相见恨晚。

4. int half(int *array,int len,int key)
{
int l=0,r=len;
while(l<r)
{
int m=(l+r)>>1;
if(key>array )l=m+1;
else if(key<array )r=m;
else return m;
}
return -1;
}
这种就能避免一些Bug
l,m,r
左边是l,m;右边就是m+1,r;

5. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。