生成所有可能的子集的 C++ 程序，每个子集中的元素恰好为 k

2023-06-16 17:36:04 80

算法

Begin function PossibleSubSet(char a[], int reqLen, int s, int currLen, bool check[], int l): If currLen > reqLen Return Else if currLen = reqLen Then print the new generated sequence. If s = l Then return no further element is left. For every index there are two options: either proceed with a start as ‘true’ and recursively call PossibleSubSet() with incremented value of ‘currLen’ and ‘s’. Or proceed with a start as ‘false’ and recursively call PossibleSubSet() with only incremented value of ‘s’. End

示例

#include using namespace std; void PossibleSubSet(char a[], int reqLen, int s, int currLen, bool check[], int l) //打印给定数组集的所有可能组合 { if(currLen > reqLen) return; else if (currLen == reqLen) { cout<<"\t"; for (int i = 0; i < l; i++) { if (check[i] == true) { cout<>n; char a[n]; cout<<"\n"; for(i = 0; i < n; i++) { cout<<"Enter "<>a[i]; check[i] = false; } cout<<"\nEnter the length of the subsets required: "; cin>>m; cout<<"\nThe possible combination of length "<输出结果

输入元素数量： 7
Enter 1 element: 7
Enter 2 element: 6
Enter 3 element: 5
Enter 4 element: 4
Enter 5 element: 3
Enter 6 element: 2
Enter 7 element: 1
Enter the length of the subsets required: 6
The possible combination of length 6 for the given array set:
7 6 5 4 3 2
7 6 5 4 3 1
7 6 5 4 2 1
7 6 5 3 2 1
7 6 4 3 2 1
7 5 4 3 2 1
6 5 4 3 2 1

生成所有可能的子集的 C++ 程序，每个子集中的元素恰好为 k

算法

示例

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