Этот пост дополняет "What is interesting in the work of E.A.Mironchick "Solving problems EGE-18 with per bit operations "
(x&248 ¬= 152) v (x&252 = 156) v (x&250 = 154) v (x&A = 0) = 1
252 = 11111100
156 = 10011100
248 = 11111000
152 = 10011000
250 = 11111010
154 = 10011010
Согласно формулам К.Ю. Полякова
(x&248 ¬= 152) = ¬Z(96) + Z(128) + Z(16) + Z(8)
(x&252 = 156) = Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
(x&250 = 154) = Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2)
¬Z(96) + Z(128) + Z(16) + Z(8) + Z(64)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
+ Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2) + A = 1
¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4)
+ Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2) + A = 1
¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4) + ¬Z(2) + A = 1
¬(Z(96)*Z(4)*Z(2)) + Z(128) + Z(16) + Z(8) + A = 1
Z(102) => Z(128) + Z(16) + Z(8) + A = 1
(Z(102)=>Z(128)) + (Z(102)=>Z(16)) + (Z(102)=>Z(8)) + (Z(102)=>A) = 1
Z(102) => A = 1
Тогда A(max) = 102
Детально здесь
https://informatics-ege.blogspot.com/2018/12/bitwise2-extended-functionality.html