COCI '15竞赛6 #4 Parovi

def count_ones(num):
    count = 0
    while num:
        count += num & 1
        num >>= 1
    return count

def find_pairs(A):
    pairs = {}
    for i in range(len(A)):
        for j in range(i+1, len(A)):
            if count_ones(A[i]) == count_ones(A[j]):
                key = count_ones(A[i])
                if key not in pairs:
                    pairs[key] = []
                pairs[key].append((A[i], A[j]))
    return pairs

# 测试示例
A = [1, 2, 3, 4, 5]
result = find_pairs(A)
for key, value in result.items():
    print("1的个数为{}的元素对: {}".format(key, value))

1的个数为1的元素对: [(1, 2), (1, 4), (2, 4)]
1的个数为2的元素对: [(3, 5)]