Python 计算24点

import operator from itertools import product,permutations def mydiv(n,d): return operator.truediv(n,d) if d!=0 else 999999#不能直接用除法，是因为可能除0报错了，所以要设定这个 sys=[operator.add,operator.sub,operator.mul,mydiv] op={sy:b for sy,b in zip(sys,"+-*/")}#zip,作用是为了一一对应上 def s24(num): for x,y,z in product(sys,repeat=3):#可以重复的排列组合，比如aaa，bbb for a,b,c,d in permutations(num):#（主要是组合，不能重复，比如abcd，cbad） if x(y(a,b),z(c,d))==24: return f"({a}{op[y]}{b}){op[x]}({c}{op[z]}{d})" elif x(y(z(a,b),c),d)==24: return f"(({a}{op[z]}{b}){op[y]}{c}){op[x]}{d}" elif x(y(a,z(b,c)),d)==24: return f"({a}{op[y]}({b}{op[z]}{c})){op[x]}{d}" elif x(a,y(b,z(c,d)))==24: return f"{a}{op[y]}({b}{op[y]}({c}{op[z]}{d}))" return "no answer"#这个要等全部遍历完才输出，不然一个遍历到中间输出了就停止计算了 for nums in [ [2, 2, 2, 9], [7, 9, 2, 8], [7, 7, 1, 2], [10, 10, 4, 4], [9, 9, 10, 6], [5, 5, 1, 5], [5, 5, 2, 10], [4, 1, 5, 6], [7, 3, 3, 7], [8, 3, 8, 3]]: print(f"s24({nums}) -> {s24(nums)}")