续"如何捅破python编程的那层纸之三"，,# coding:utf

续"如何捅破python编程的那层纸之三"，,# coding:utf

# coding:utf-8import itertoolsdef sameSums1(int_list):    if len(int_list) == 1:        # 列表只有一个元素必不能等分两组        return False    sum_of_lsit = sum(int_list)    halfsum = sum_of_lsit / 2    if sum_of_lsit % 2:        # 和为奇数不可等分        return False    int_list.sort(reverse=True)    for i in xrange(1, len(int_list)/2 + 1):        if sum(int_list[:i]) < halfsum:            # 若最大的前N位之和不足半值就不必检测N位组合的各种方案了            continue        elif sum(int_list[:i]) == halfsum:            # 若恰巧是半值就直接返回            return tuple(int_list[:i])        for subset in itertools.combinations(int_list, i):            sumsub = sum(subset)            if sumsub == halfsum:                # 找到就结束                return subsetif __name__ == "__main__":    import doctest    doctest.testmod()    import time    lst1 = [3, 9, 10, 30, 8]    lst2 = [4, 5, 6, 7, 8]    lst3 = [2, 2, 2, 3, 3]    lst4 = [10, 9, 8, 6, 5, 3, 2, 1]    lst5 = [74, 80, 40, 38, 80, 68, 58, 91, 65, 10, 75, 88, 47, 84, 30]    lst6 = [97, 2, 87, 96, 24, 12, 98, 85, 99, 67, 49, 86, 65,            80, 83, 28, 57, 68, 90, 2, 62, 93, 9, 23]    lst7 = [93, 99, 86, 94, 36, 62, 4, 54, 88, 69, 64, 19, 21,            30, 53, 24, 10, 7, 93, 46, 82, 78, 20, 45, 27, 91,            72, 48, 90, 62, 57, 65, 49]    t0 = time.time()    print sameSums1(lst1)    print sameSums1(lst2)    print sameSums1(lst3)    print sameSums1(lst4)    print sameSums1(lst5)    print sameSums1(lst6)    print sameSums1(lst7)    t1 = time.time()    print t1 - t0