大多数平衡二分的置换

Rory的评论给了我正确的方向:

• 从开始 0...01...1 第一个 k/2 一个,然后是 k/2 - 1 , k/2 - 2 等等

• Gosper's Hack

一个简单的实现可能如下所示(对于 k <= 63

for (int i = k / 2; i > 0; --i) {
    // start with 0 ... 0 1 ... 1 (i times)
    unsigned int v = (1 << i) - 1;
    // first bitset that doesn't represent a valid bipartition
    unsigned int end = 1 << k;
    // without this, we would count some bipartitions twice for k even
    if (k % 2 == 0 && i == k / 2) end >>= 1;
    while(v < end) {
        // do something with v...
        // iterate to the lexicographically next permutation
        unsigned int t = v | (v - 1);
        v = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
    }
}