问个傻问题，24点出牌总共有几种组合？

cat · #1 (**permalink**) 2006-06-28

是不是
for (i = 0; i < 13; ++i)
for (j = i; j < 13; ++j)
for (k = j; k < 13; ++k)
for (l = k; l < 13; ++l)
++count;
的数目？
不过想直接计算就不会了……

polyrandom · #2 (**permalink**) 2006-06-28

1234和4321应该不算一种组合吧？
13^4应该是不对的。

zweily · #3 (**permalink**) 2006-06-28

1234和4321是一样的阿。这个就是组合阿，不是排列阿。
所以么，就是52张牌取4张牌的排列嘛

cat · #4 (**permalink**) 2006-06-29

52张取4张的组合也不对
黑桃A 黑桃2 黑桃3 黑桃4
和
草花A 草花2 草花3 草花4
是同一种组合

to pora：
1234和4321认为是一样的吧。你算24点的时候可以任意组合的。

polyrandom · #5 (**permalink**) 2006-06-29

sorry，我笔误了。我本来想说的就是，因为1234和4321是相同的，所以这样的4重循环肯定是不对的

排列组合问题

Juvexp · #6 (**permalink**) 2006-06-29

C(x,y)代表从x张中选y张的组合

C(13,4) // 4张都不一样的
+C(13,3) // 仅有两张牌一样的
+C(13,2) // 仅有三张牌一样的
+C(13,1) // 4张都一样的

是不是这个

zero · #7 (**permalink**) 2006-06-29

I think the answer of Juvexp is correct.
To cat: I have written a program to verify the answer. The algorithm is the most naive one. I think it is correct.

I think C(52, 4) is also correct, if we regard 4C and 4H is different. But I think since we get the same result from 4 no matter it is 4C or 4H, maybe neglecting the difference is better

cat · #8 (**permalink**) 2006-06-29

引用:

作者: polyrandom

sorry，我笔误了。我本来想说的就是，因为1234和4321是相同的，所以这样的4重循环肯定是不对的

排列组合问题

注意内层循环的起始不是0.

to zero:
I don't think so.
For example, 2 2 3 3 is a legal combination, which is not counted in his formula. However this is a big hint, made me to remember something of the permutations and combinations

PS, would you kindly paste your code here?

zero · #9 (**permalink**) 2006-06-29

...
我没希望了。
Juvexp的那个方法，把(2, 3, 3, 3)和(2, 2, 3, 3)看作一种了。
这样看来，我那个code也是错的了，

我把四个数的比较用Contains了。

以前看的Combinatorics全忘光了。

tomato · #10 (**permalink**) 2006-06-29

P(13, 4) / P(4, 4) --- 1 1 1 1
+ P(13, 1) * P(12, 2) / P(2, 2) --- 2 1 1
+ P(13, 1) * P(12, 1) --- 3 1
+ P(13 2) / P(2 2) --- 2 2

对于这种可重复的组合问题
有没有除穷举外的方法？
比如有n种物品，每种有a_i个( 1 <= i <= n )
取m个物品有几种取法

polyrandom · #11 (**permalink**) 2006-06-30

引用:

作者: cat

注意内层循环的起始不是0.

to zero:
I don't think so.
For example, 2 2 3 3 is a legal combination, which is not counted in his formula. However this is a big hint, made me to remember something of the permutations and combinations

PS, would you kindly paste your code here?

我再脸红一下

polyrandom · #12 (**permalink**) 2006-06-30

正确算法应该是：
C(13, 4 ) = 715
C(13, 3 ) * 3 = 858
C(13, 2 ) * 3 = 234
C(13, 1 ) = 13

验证程序：

代码:

#include <iostream>
#include <set>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
	set<vector<int> > cardSets;
	for( int i = 0; i < 13; ++i )
		for( int j = 0; j < 13; ++j )
			for( int k = 0; k < 13; ++k )
				for( int l = 0; l < 13; ++l )
				{
					int cards[] = { i, j, k, l };
					std::sort( cards, cards + sizeof( cards ) / sizeof( *cards ) );
					cardSets.insert( vector<int>( cards, cards + sizeof( cards ) / sizeof( *cards ) ) );
				}
	cout << cardSets.size() << endl;
}

Juvexp · #13 (**permalink**) 2006-06-30

引用:

作者: tomato

P(13, 4) / P(4, 4) --- 1 1 1 1
+ P(13, 1) * P(12, 2) / P(2, 2) --- 2 1 1
+ P(13, 1) * P(12, 1) --- 3 1
+ P(13 2) / P(2 2) --- 2 2

对于这种可重复的组合问题
有没有除穷举外的方法？
比如有n种物品，每种有a_i个( 1 <= i <= n )
取m个物品有几种取法

还应该加上4张都一样的：P(13,1)