题
假设我给出:
- 一个范围的整数
iRange
(即从1
起来iRange
)和 - 所需数量的组合
我想找到的所有数量的可能的组合,并打印出所有这些的组合。
例如:
给予: iRange = 5
和 n = 3
然后组合的数量是 iRange! / ((iRange!-n!)*n!) = 5! / (5-3)! * 3! = 10
组合,并且输出为:
123 - 124 - 125 - 134 - 135 - 145 - 234 - 235 - 245 - 345
另一个例子:
给予: iRange = 4
和 n = 2
然后组合的数量是 iRange! / ((iRange!-n!)*n!) = 4! / (4-2)! * 2! = 6
组合,并且输出为:
12 - 13 - 14 - 23 - 24 - 34
我尝试迄今为止是:
#include <iostream>
using namespace std;
int iRange= 0;
int iN=0;
int fact(int n)
{
if ( n<1)
return 1;
else
return fact(n-1)*n;
}
void print_combinations(int n, int iMxM)
{
int iBigSetFact=fact(iMxM);
int iDiffFact=fact(iMxM-n);
int iSmallSetFact=fact(n);
int iNoTotComb = (iBigSetFact/(iDiffFact*iSmallSetFact));
cout<<"The number of possible combinations is: "<<iNoTotComb<<endl;
cout<<" and these combinations are the following: "<<endl;
int i, j, k;
for (i = 0; i < iMxM - 1; i++)
{
for (j = i + 1; j < iMxM ; j++)
{
//for (k = j + 1; k < iMxM; k++)
cout<<i+1<<j+1<<endl;
}
}
}
int main()
{
cout<<"Please give the range (max) within which the combinations are to be found: "<<endl;
cin>>iRange;
cout<<"Please give the desired number of combinations: "<<endl;
cin>>iN;
print_combinations(iN,iRange);
return 0;
}
我的问题:
我的代码相关的印刷的合作仅用于 n = 2, iRange = 4
我不能让它工作在一般情况下,即对任何 n
和 iRange
.
解决方案
这是你的代码行编辑:D:D 递归 方案:
#include <iostream>
int iRange=0;
int iN=0; //Number of items taken from iRange, for which u want to print out the combinations
int iTotalCombs=0;
int* pTheRange;
int* pTempRange;
int find_factorial(int n)
{
if ( n<1)
return 1;
else
return find_factorial(n-1)*n;
}
//--->Here is another solution:
void print_out_combinations(int *P, int K, int n_i)
{
if (K == 0)
{
for (int j =iN;j>0;j--)
std::cout<<P[j]<<" ";
std::cout<<std::endl;
}
else
for (int i = n_i; i < iRange; i++)
{
P[K] = pTheRange[i];
print_out_combinations(P, K-1, i+1);
}
}
//Here ends the solution...
int main()
{
std::cout<<"Give the set of items -iRange- = ";
std::cin>>iRange;
std::cout<<"Give the items # -iN- of iRange for which the combinations will be created = ";
std::cin>>iN;
pTheRange = new int[iRange];
for (int i = 0;i<iRange;i++)
{
pTheRange[i]=i+1;
}
pTempRange = new int[iN];
iTotalCombs = (find_factorial(iRange)/(find_factorial(iRange-iN)*find_factorial(iN)));
std::cout<<"The number of possible combinations is: "<<iTotalCombs<<std::endl;
std::cout<<"i.e.the combinations of "<<iN<<" elements drawn from a set of size "<<iRange<<" are: "<<std::endl;
print_out_combinations(pTempRange, iN, 0);
return 0;
}
其他提示
你的解决方案将永远只能工作n=2。考虑使用一系列(梳)与n整数,再循环将会勾起来的最后一项数。当,项目达到最大的更新,然后梳[n-2]项和设定的最后一个项目前的价值+1.
基本上工作就像一个时钟,但你需要逻辑找到什么要提升和什么未来的最低值。
看起来像一个很好的问题递归。
定义功能 f(prefix, iMin, iMax, n)
, 那打印的所有组合 n
数字的范围[iMin
, iMax
]和返回总数的组合。对于 n
=1,它应该打印的每一位从 iMin
要 iMax
和返回 iMax - iMin + 1
.
你的 iRange = 5
和 n = 3
种情况下,你的呼叫 f("", 1, 5, 3)
.输出应 123 - 124 - 125 - 134 - 135 - 145 - 234 - 235 - 245 - 345
.
注意到第一组产出的简单 1
作为前缀到的产出 f("", 2, 5, 2)
, 即 f("1", 2, 5, 2)
, ,随后通过 f("2", 3, 5, 2)
和 f("3", 4, 5, 2)
.看看你会怎么做,用一个循环。此之间,情况 n
=1所述,和陷阱不良的投入(最好的,如果他们打印什么和返回0,它应该简化循环)中,应该能够写 f()
.
我停止短,因为这看起来像一个家庭作业。这足以让你开始的?
编辑:只是为了笑声,我写了一蟒蛇的版本。蟒蛇有一个容易的时候周围投掷集和列表的事情,并保持清晰可见。
#!/usr/bin/env python
def Combos(items, n):
if n <= 0 or len(items) == 0:
return []
if n == 1:
return [[x] for x in items]
result = []
for k in range(len(items) - n + 1):
for s in Combos(items[k+1:], n - 1):
result.append([items[k]] + s)
return result
comb = Combos([str(x) for x in range(1, 6)], 3)
print len(comb), " - ".join(["".join(c) for c in comb])
注意, Combos()
不关心这种类型的项目 items
名单。
这里有一个例子的一个普通的递归的解决方案。我相信存在一个更多的最佳方式实现的,如果你替换用递归的循环。这可能是你的家庭作业)
#include <stdio.h>
const int iRange = 9;
const int n = 4;
// A more efficient way to calculate binomial coefficient, in my opinion
int Cnm(int n, int m)
{
int i;
int result = 1;
for (i = m + 1; i <= n; ++i)
result *= i;
for (i = n - m; i > 1; --i)
result /= i;
return result;
}
print_digits(int *digits)
{
int i;
for (i = 0; i < n; ++i) {
printf("%d", digits[i]);
}
printf("\n");
}
void plus_one(int *digits, int index)
{
int i;
// Increment current digit
++digits[index];
// If it is the leftmost digit, run to the right, setup all the others
if (index == 0) {
for (i = 1; i < n; ++i)
digits[i] = digits[i-1] + 1;
}
// step back by one digit recursively
else if (digits[index] > iRange) {
plus_one(digits, index - 1);
}
// otherwise run to the right, setting up other digits, and break the recursion once a digit exceeds iRange
else {
for (i = index + 1; i < n; ++i) {
digits[i] = digits[i-1] + 1;
if (digits[i] > iRange) {
plus_one(digits, i - 1);
break;
}
}
}
}
int main()
{
int i;
int digits[n];
for (i = 0; i < n; ++i) {
digits[i] = i + 1;
}
printf("%d\n\n", Cnm(iRange, n));
// *** This loop has been updated ***
while (digits[0] <= iRange - n + 1) {
print_digits(digits);
plus_one(digits, n - 1);
}
return 0;
}
这是我的C++的功能,与不同的接口(根据sts::set),但执行同一任务:
typedef std::set<int> NumbersSet;
typedef std::set<NumbersSet> CombinationsSet;
CombinationsSet MakeCombinations(const NumbersSet& numbers, int count)
{
CombinationsSet result;
if (!count) throw std::exception();
if (count == numbers.size())
{
result.insert(NumbersSet(numbers.begin(), numbers.end()));
return result;
}
// combinations with 1 element
if (!(count - 1) || (numbers.size() <= 1))
{
for (auto number = numbers.begin(); number != numbers.end(); ++number)
{
NumbersSet single_combination;
single_combination.insert(*number);
result.insert(single_combination);
}
return result;
}
// Combinations with (count - 1) without current number
int first_num = *numbers.begin();
NumbersSet truncated_numbers = numbers;
truncated_numbers.erase(first_num);
CombinationsSet subcombinations = MakeCombinations(truncated_numbers, count - 1);
for (auto subcombination = subcombinations.begin(); subcombination != subcombinations.end(); ++subcombination)
{
NumbersSet cmb = *subcombination;
// Add current number
cmb.insert(first_num);
result.insert(cmb);
}
// Combinations with (count) without current number
subcombinations = MakeCombinations(truncated_numbers, count);
result.insert(subcombinations.begin(), subcombinations.end());
return result;
}