The calculation process is as follows: Look at A first, and A can only be 1 or 2. If it is more than 3, multiply it by 4, which means that the number becomes 6 digits instead of 5 digits.

Starting from 1, if it is 1, then the e of this number should be 4, or a number above 4 (plus the previous decimal places). And numbers above 4, that is, e multiplied by 4, will not get 1. So A can't be 1, only 2.

If a is 2, then e is 8 or above. But if 9 is multiplied by 4, several digits are 6, which is different from the previous result that A is 2. So e can only be 8.

Look at b again, because e is 8, that is to say, there is no decimal place in the first place, so b can only be 1 or 2, and it will not be carried after being multiplied by 4. A=2, so b= 1.

Look at D, because B is 1, E is 8, E is multiplied by 4, followed by 3, 3 plus 8, 18, 28. . . Equal numbers can get 1 of b, so d may be 2 or 7. A=2, so d is 7.

Look at C.B.' s 1 multiplied by 4, plus 3 equals 7 of D. So after C is multiplied by 4, plus the previous D multiplied by 4+3, it must be 3. It can only be a number of 7 or above. Seven, eight already have. So c can only be 9.

Second, first of all, if A can't be 1 and 2, 1, it will definitely not work. The following result is abcab2. If you want to be 6, B can only be 5-9. None of these attempts will work.

Try from 3. Suppose A is 3, multiply A by 10 to get 9, and B can no longer be 3, but if B is multiplied by 3 and then goes forward by 1, D becomes 0, which is wrong; So I can only go forward 2. In other words, B is 7-9. 7, which is exactly 2 1, is also 1 1 after entering 2. It seems ok. Let's keep trying.

If the number of the preceding c is 1, it can only be 0, and the preceding 9+2 decimal 1 is used. Then go forward, B is 7 times 3 to get 2 1, A times 9 of A, and 2 digits are added to get 1 1. . Exactly 3 is right.

So the answer is: a, 3 b, 7 c, 0 d, 1