1 /*
2 题意：找出一个0和1组成的数字能整除n
3 DFS：200的范围内不会爆long long，DFS水过~
4 */
5 /************************************************
6 Author :Running_Time
7 Created Time :2015-8-2 14:21:51
8 File Name :POJ_1426.cpp
9 *************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 typedef long long ll;
31 const int MAXN = 1e4 + 10;
32 const int INF = 0x3f3f3f3f;
33 const int MOD = 1e9 + 7;
34 ll n;
35 bool ok;
36
37 void DFS(ll x, int step, int dep) {
38 if (ok) return ;
39 if (step >= dep) return ;
40 if (x % n == 0) {
41 printf ("%I64d
", x);
42 ok = true; return ;
43 }
44 DFS (x * 10, step + 1, dep);
45 DFS (x * 10 + 1, step + 1, dep);
46 }
47
48 int main(void) { //POJ 1426 Find The Multiple
49 while (scanf ("%I64d", &n) == 1) {
50 if (!n) break;
51 ok = false; ll dep = 1;
52 while (true) {
53 if (ok) break;
54 DFS (1, 0, dep); dep++;
55 }
56 }
57
58 return 0;
59 }
1 /*
2 BFS+同余模定理：mod数组保存，每一位的余数，当前的数字由少一位的数字递推，
3 比如4(100)可以从2(10)递推出，网上的代码太吊，膜拜之
4 详细解释：http://blog.csdn.net/lyy289065406/article/details/6647917
5 */
6 /************************************************
7 Author :Running_Time
8 Created Time :2015-8-2 15:14:33
9 File Name :POJ_1426_BFS.cpp
10 *************************************************/
11
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e6 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 int mod[MAXN];
38 int ans[MAXN];
39
40 int main(void) {
41 int n;
42 while (scanf ("%d", &n) == 1) {
43 if (!n) break;
44 mod[1] = 1; int i;
45 for (i=2; mod[i-1]; ++i) {
46 mod[i] = (mod[i/2] * 10 + (i&1)) % n; //BFS双入口模拟
47 }
48 int t = 0; i--;
49 while (i) {
50 ans[++t] = i & 1;
51 i /= 2;
52 }
53 while (t) {
54 printf ("%d", ans[t--]);
55 }
56 puts ("");
57 }
58
59 return 0;
60 }