DFS/BFS(同余模) POJ 1426 Find The Multiple

题目传送门

 /*
     题意：找出一个0和1组成的数字能整除n
     DFS：200的范围内不会爆long long，DFS水过~
 */
 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-2 14:21:51
 File Name     :POJ_1426.cpp
 *************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e4 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 ll n;
 bool ok;

 void DFS(ll x, int step, int dep)  {
     if (ok) return ;
     if (step >= dep)    return ;
     if (x % n == ) {
         printf ("%I64d\n", x);
         ok = true;  return ;
     }
     DFS (x * , step + , dep);
     DFS (x *  + , step + , dep);
 }

 int main(void)    {       //POJ 1426 Find The Multiple
     while (scanf ("%I64d", &n) == )   {
         if (!n) break;
         ok = false; ll dep = ;
         while (true)    {
             if (ok) break;
             DFS (, , dep);    dep++;
         }
     }

     return ;
 }

 /*
     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，
         比如4(100)可以从2(10)递推出，网上的代码太吊，膜拜之
     详细解释：http://blog.csdn.net/lyy289065406/article/details/6647917
 */
 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-2 15:14:33
 File Name     :POJ_1426_BFS.cpp
 *************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int mod[MAXN];
 int ans[MAXN];

 int main(void)    {
     int n;
     while (scanf ("%d", &n) == )   {
         if (!n) break;
         mod[] = ; int i;
         for (i=; mod[i-]; ++i)    {
             mod[i] = (mod[i/] *  + (i&)) % n;       //BFS双入口模拟
         }
         int t = ;  i--;
         while (i)   {
             ans[++t] = i & ;
             i /= ;
         }
         while (t)   {
             printf ("%d", ans[t--]);
         }
         puts ("");
     }

     return ;
 }

BFS