Acdream 1417 Numbers (Violent enumeration)

Transmission Door
Numbers
Time limit:2000/1000ms (java/others) Memory limit:128000/64000kb (java/others)
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Problem Description

  Consider numbers from 1 to n.  You have to find the smallest lexicographically number among them which is divisible by k.

Input

  Input file contains several test cases. Each test case consists of two integer numbers n and k on a line(1 ≤ n ≤ 1018, 1 ≤ k ≤ n).  The last test case is followed by a line that contains two zeroes. This line must not be processed.

Output

  For each test case output one integer number — the smallest lexicographically number not exceeding n which is divisible by k.

Sample Input

2000 17
2000 20
2000 22
0 0
Sample Output

1003
100
1012
Hint

Multiple sets of data
Source

Andrew Stankevich Contest 22

Main topic:
give you a number n (n<10^18), which allows you to ask for the smallest of the least N of the dictionary order to be divisible by K
Problem Solving Ideas:
Need to note here is the dictionary order that is 1001:123 small, in fact, we can enumerate 10^p (p<=18), then up to 18 bits, we put 10^p after the first to K modulo equals 0 of the number of digital groups recorded (note that the enumeration does not exceed N), The final answer must be in here, and we'll just have to convert it to a string to do the dictionary sequence.

My Code:

#include <iostream>#include <cstring>#include <algorithm>Using namespace Std;typedefLong LongLL;Char Str[ -][ -];string s[ -]; LL a[ -];intMain () {LL n, K; while(cin>>n>>k) {if(!n &&!k) Break;intp =0; LL TP = n; while(TP)            {p++; TP/=Ten; } TP =1; for(intI=0; i<p-1; i++) TP *=Ten;intCNT =0; LL tmp = tp%k;if(TMP) TMP = TP+K-TMP;ElseTMP = TP;if(TMP <= N) a[cnt++] = tmp; TP/=Ten; while(TP) {tmp = Tp%k;if(TMP) TMP = TP+K-TMP;ElseTMP = TP;if(TMP < tp*Ten) a[cnt++] = tmp; TP/=Ten; } for(intI=0; i<cnt; i++) {int sum=0; while(A[i]) {intTP = a[i]%Ten;StrI [sum+ +] = tp+' 0 '; A[i]/=Ten; }StrI [sum] =' + '; } for(intI=0; i<cnt; i++) {intLen = strlen (Str[i]);///cout<<str[i]<<endl;             for(intj=0; j<len/2; J + +) {CharCH =StrI [len-j-1];StrI [len-j-1] =StrI [j];StrI            [j] = ch; }        } for(intI=0; i<cnt; i++) S[i] =Str[i];        Sort (s, s+cnt); cout<<s[0]<<endl; }return 0;}

Acdream 1417 Numbers (Violent enumeration)