C++ 如何确保n位中至少有m位处于启用状态？

我有n个集合，需要确保至少选择了m个集合。我打算处理比特。 我的做法是：

for i in [0,(2^n)-1]
   convert i to binary
       if number of 1s are greater than or equal to m
            { Some calculations requiring which bits are on }

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现在，有没有其他方法可以确保on位的数量至少为m？在我上面的方法中，我将浪费时间将数字转换为二进制，然后检查on位的数量是否>=m。有没有办法缩短循环？我用的是C++

我想您需要生成位掩码，以便从一组n个元素中选择至少m个元素的子集。 如果我们有一个算法来生成正好有m个位集的所有位掩码，那么这很容易做到

#include <iostream>
#include <vector>
#include <cassert>
using namespace std;

// Given "n" and "r", generate all the possible nCr subsets of an array of size "n"

typedef unsigned long long ULL;

// Generate the lowest number bigger than "num" having exactly "r" set bits
// Algorithm (From Cracking The Coding Interview, 5th Edition) -:
//  1.  Find the position of the rightmost "non-trailing" zero (such that there is atleast one '1' on its right).
//          Let this position be 'p'
//          If there does not exist such a zero, the input is already the largest number possible.
//  2.  Flip the bit at 'p' to 1 from 0.
//  3.  Count the number of zeroes and ones to the right of 'p'.
//          Let that number be c0 and c1 respectively.
//  4.  Set all the bits to the right of 'p' to 0.
//  5.  Set the first (c1-1) bits (starting from the right) to 1.
ULL NextBigger( ULL num )
{
    ULL numBak = num;

    // Compute c0 and c1
    // c0 = Number of zeroes to the right of the rightmost non-trailing zero
    size_t c0 = 0;
    // c1 = Number of ones to the right of the rightmost non-trailing zero
    size_t c1 = 0;
    while ( numBak && ( ( numBak & 1 ) == 0 ) )
    {
        c0++;
        numBak >>= 1;
    }
    while ( ( numBak & 1 ) == 1 )
    {
        c1++;
        numBak >>= 1;
    }

    // If the input is either 0,
    // or of the form "1111..00000",
    // then there is no bigger number possible
    // Note that for this to work, num should be unsigned
    if ( c0 + c1 == 0 || c0 + c1 == ( sizeof( num ) * 8 ) )
    {
        return 0;
    }

    // Position of the rightmost non-trailing zero ( starting from the right )
    const size_t p = c0 + c1;

    // Flip the rightmost non-trailing zero
    num |= 1 << p;
    // Clear all bits to the right of p
    num &= ~( ( 1 << p ) - 1 );
    // Insert (c1-1) ones on the right of p
    num |= ( 1 << ( c1 - 1 ) ) - 1;

    return num;
}

vector<ULL> GenerateSubsets( const size_t& n, const size_t& r )
{
    assert( n > 0 );
    assert( r > 0 );
    assert( n >= r );

    vector<ULL> subsets;

    // The smallest number having exactly "r" bits set
    ULL lowest = ( 1ULL << r ) - 1;
    // The biggest number having exactly "r" bits set
    ULL highest = lowest << ( n - r );

    // The set bits in the binary of "bitMask" denote the positions of the set included in the subset
    // This loop should run exactly nCr times
    for ( ULL bitMask = lowest; bitMask <= highest; bitMask = NextBigger( bitMask ) )
    {
        subsets.push_back( bitMask );
    }

    return subsets;
}

// Extracts the subset indices from the bitmask
vector<size_t> DecodeMask( ULL bitMask )
{
    vector<size_t> positions;

    size_t i = 0;

    while ( bitMask )
    {
        if ( bitMask & 1 )
        {
            positions.push_back( i );
        }
        bitMask >>= 1;
        i++;
    }

    return positions;
}

int main()
{
    size_t n = 5;
    size_t r = 2;

    cout << "Generating subsets of size " << r << "\n";

    auto vec = GenerateSubsets( n, r );
    cout << "Number of subsets = " << vec.size() << "\n";

    // Print the subset indices
    for ( size_t i = 0; i < vec.size(); i++ )
    {
        auto decode = DecodeMask( vec[i] );
        for ( size_t j = 0; j < decode.size(); j++ )
        {
            cout << decode[j] << " ";
        }
        cout << "\n";
    }
}

现在我们可以很容易地修改它，通过应用上述算法，在将m增加到n的同时，生成至少设置了m位的所有位掩码

// Rest of the code same as above

int main()
{
    size_t n = 5;
    size_t m = 2;

    for ( size_t r = m; r <= n; r++ )
    {
        cout << "Generating subsets of size " << r << "\n";
        auto vec = GenerateSubsets( n, r );
        cout << "Number of subsets = " << vec.size() << "\n";

        // Print the subset indices
        for ( size_t i = 0; i < vec.size(); i++ )
        {
            auto decode = DecodeMask( vec[i] );
            for ( size_t j = 0; j < decode.size(); j++ )
            {
                cout << decode[j] << " ";
            }
            cout << "\n";
        }
        cout << "\n";
    }
}

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使用IMHO可以最容易地实现这一点。在大小为n的位集中，如何实现至少m位被设置为1的事实？只需使用索引运算符循环，并计算有多少位产生true。@πάνταῥεῖ 对不起，我不能得到它，如果有一个位集，比如说3位，那么有8种可能性，如何生成所有的可能性？如果m是2，如何确保至少2位设置为true，而不生成0或1位打开的可能性？是否要检查在某些现有数字中是否至少m位为1，或者是否要生成满足该条件的所有数字？