问题
I'm trying to solve this problem on SPOJ INUMBER.
Problem statement is as follows:
For the given number n find the minimal positive integer divisable by n, with the sum of digits equal to n.
INPUT
t – the number of test cases, then t test cases follow. (t <= 50)
Test case description: n - integer such that 0 < n <= 1000
OUTPUT
For each test case output the required number (without leading zeros).
EXAMPLE:
Input:
2
1
10
Output:
1
190
I can only think of a brute force solution emulating the number digit by digit from 0-9 and forming a dfs structure and repeatedly checking whether it's divisible by n or not.
Before asking my question here, I did a meticulous search on this problem on the internet and couldn't found any detailed explanation. Most of them were undocumented raw code and others were giving just a jist of the solution.
I'm really interested in solving this problem not just for the points but to learn something new.
Thanks for the help Stackoverflow community :)
回答1:
You can use a breadth first search.
Let num(p, q) be the minimum number of digits with digit sum p and remainder mod n equal to q.
We want to find num(n, 0). Then, we can greedily build the smallest such number.
We start from the state (0, 0). From a state (x, y) you can get to a state:
(x + j, (y * 10 + j) % n)
for each digit j.
Keep track of each digit j you add and then backtrack from (n, 0) to (0, 0).
There are some implementation details to figure out. If you get stuck, I have found some implementations online: on topcoder and on github.
来源：https://stackoverflow.com/questions/32426222/spoj-inumber-cant-seem-to-develop-a-solution-within-the-time-limit