5. Простой линейный алгоритм для формального исполнителя

(‘2’, ‘1’, ‘1’, ‘1’, ‘1’)

import itertools
sp = itertools.product('012',repeat = 5)
for t in sp:
    r = 65
    for k in t:
        if k == '1':
            r = r / 2
        if k == '2':
            r = r - 1
    if r == 4 and t.count('0')< 3:
        print(t)

(‘1’, ‘1’, ‘2’, ‘2’)

import itertools
sp = itertools.product('012',repeat = 4)
for t in sp:
    r = 3
    for k in t:
        if k == '1':
            r = r ** 2
        if k == '2':
            r = r + 3
    if r == 87 and t.count('0')< 3:
        print(t)

(‘2’, ‘1’, ‘1’, ‘2’)

import itertools
sp = itertools.product('012',repeat = 4)
for t in sp:
    r = 4
    for k in t:
        if k == '1':
            r = r * 10 + 3
        if k == '2':
            r = r + 4
    if r == 837 and t.count('0')< 3:
        print(t)

4

for x in range(2,44):
    b = 3
    #12112
    b = b + x #1
    b = b * 2 #2
    b = b + x #1
    b = b + x #1
    b = b * 2 #2
    if b == 44:
        print(x)

40

for x in range(2,100):
    b = 5
    #21222
    b = b - 4 #2
    b = b * x #1
    b = b - 4 #2
    b = b - 4 #2
    b = b - 4 #2
    if b == 28:
        print(x)

20

for x in range(2,160):
    b = 100
    #11122
    b = b - x #1
    b = b - x #1
    b = b - x #1
    b = b * 2 #2
    b = b * 2 #2
    if b == 160:
        print(x)

3

for b in range(2,17):
    x = 2
    #21121
    x = x * 5 #2
    x = x - b #1
    x = x - b #1
    x = x * 5 #2
    x = x - b #1
    if x == 17:
        print(b)

3

for x in range(2,48):
    b = 6
    #21313
    b = b - 2 #2
    b = b * 3 #1
    b = b + x #3
    b = b * 3 #1
    b = b + x #3
    if b == 48:
        print(x)

2

for k in range(0,328):
    b = 4
    #12122
    b = b ** 2 #1
    b = b + k #2
    b = b ** 2 #1
    b = b + k #2
    b = b + k #2
    if b == 328:
        print(k)

8

for x in range(2,528):
    b = 528
    #21221
    b = b - 16  #2
    b = b / x   #1
    b = b - 16  #2
    b = b - 16   #2
    b = b / x   #1
    if b == 4:
        print(x)