关于n

Giottoil

0<n<10
n#n#n=1(#可以是随机的运算符号，此处两个＃可以不一样)
求n

shuofxz

1 + 1 - 1
1 - 1 + 1
1 * 1 * 1
1 * 1 / 1
1 / 1 * 1
1 / 1 / 1

代码写的比较糙，仅做参考。。。

def get_res(cal, a, b):
    if cal == 0:
        return a+b
    elif cal == 1:
        return a-b
    elif cal == 2:
        return a*b
    elif cal == 3:
        return a/b

def print_line(num, pre_c, after_c):
    cal = ['+', '-', '*', '/']
    print(num, cal[pre_c], num, cal[after_c], num)

for i in range(1,10):
    for pre_c in range(0,4):
        pre_num = get_res(pre_c, i, i)
        for after_c in range(0,4):
            after_num = get_res(after_c, pre_num, i)
            if after_num == 1:
                print_line(i, pre_c, after_c)

shuofxz

上面的程序有问题，忘了考虑优先级的问题了=。=  改过之后多了一组答案

def cal_num(cal, a, b):
    if cal == 0:
        return a + b
    elif cal == 1:
        return a - b
    elif cal == 2:
        return a * b
    elif cal == 3:
        return a / b

def get_res(num, pre_c, after_c):
    if after_c > 1:
        result_pre = cal_num(after_c, num, num)
        result = cal_num(pre_c, num, result_pre)
        return result
    else:
        result_pre = cal_num(pre_c, num, num)
        result = cal_num(after_c, result_pre, num)
        return result

def print_line(num, pre_c, after_c):
    cal = ['+', '-', '*', '/']
    print(num, cal[pre_c], num, cal[after_c], num)

for i in range(1,10):
    for pre_c in range(0,4):
        for after_c in range(0,4):
            result = get_res(i, pre_c, after_c)
            if result == 1:
                print_line(i, pre_c, after_c)

1 + 1 - 1
1 - 1 + 1
1 * 1 * 1
1 * 1 / 1
1 / 1 * 1
1 / 1 / 1
2 - 2 / 2