Enumerate Combination C(k, n) in a bitset

標籤：algorithm   c++   bitset   

Suppose n<=32, we can enumerate C(k, n), with bits representing absence or presence, in the following way:

#include <iostream>#include <vector>#include <bitset>using namespace std;bitset<32> getComb(const vector<int> &comb) {bitset<32> bitcombs;for (int i=0; i<comb.size(); ++i) bitcombs.set(comb[i], true);return bitcombs;}bool nextComb(vector<int> &comb, int n) {int k = comb.size(), i = k - 1;++comb[i];while (i>=1 && comb[i]>=n-k+1+i) ++comb[--i];if (comb[0] > n-k) return false; for (++i; i<k; ++i) comb[i] = comb[i-1] + 1;return true;}int main() {int n = 3, k = 2;vector<int> comb(k);for (int i=0; i<k; ++i) comb[i] = i;do {cout<<getComb(comb).to_ulong()<<endl;} while (nextComb(comb, n));return 0;}

The algorithms works like this when k=4, n=5:

01111->10111->11011->11101->11110

So, could you find the pattern?