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法律研究生成每个子集恰好包含 k 个元素的所有子集的 C++ 程序
这是一个生成每个子集恰好包含 k 个元素的所有子集的 C++ 程序。
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<iostream>
using namespace std;
void PossibleSubSet(char a[], int reqLen, int s, int currLen, bool check[], int l)
//print the all possible combination of given array set
{
if(currLen > reqLen)
return;
else if (currLen == reqLen) {
cout<<"\t";
for (int i = 0; i < l; i++) {
if (check[i] == true) {
cout<<a[i]<<" ";
}
}
cout<<"\n";
return;
}
if (s == l) {
return;
}
check[s] = true;
PossibleSubSet(a, reqLen, s + 1, currLen + 1, check, l);
//recursively call PossibleSubSet() with incremented value of ‘currLen’ and ‘s’.
check[s] = false;
PossibleSubSet(a, reqLen, s + 1, currLen, check, l);
//recursively call PossibleSubSet() with only incremented value of ‘s’.
}
int main() {
int i,n,m;
bool check[n];
cout<<"Enter the number of elements: ";
cin>>n;
char a[n];
cout<<"\n";
for(i = 0; i < n; i++) {
cout<<"Enter "<<i+1<<" element: ";
cin>>a[i];
check[i] = false;
}
cout<<"\nEnter the length of the subsets required: ";
cin>>m;
cout<<"\nThe possible combination of length "<<m<<" for the given array set:\n";
PossibleSubSet(a, m, 0, 0, check, n);
return 0;
}输出
Enter the number of elements: 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
更新于：30-Jul-2019
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