kb-01-e<取余操作，宽搜，巧妙>；

题目描述：

n属于1到200，找到对应的一个数只含有0和1，并且是n的倍数；

分析：

本题有几个数会是大数；所以要考虑大数；

用到余数的性质；例如n=6，1%6=1；

1*10%6=4；              （1*10+1）%6=5；

4*10%6=4；               （4*10+1）%6=5；

5*10%6=2；                （5*10+1）%6=3；

（重复4，5）

2*10%6=2；                  。。。。=3；

3*10%6=0；

这时候发现余数为0，说明这个数可以是6的倍数；倒退回去，数分别是1，10，11，100，101，110，111，。。。。1110；

可以发现余数是一样的，同余定理；

（a*b）%n = （a%n *b%n）%n

（a+b）%n = （a%n +b%n）%n

由同余模定理  (110*10+1)%6 = (（110*10）%6+1%6 )%6 = (（110%6 * 10%6）%6 +1 )%6；

用这个同余定理就可以解决大数问题了；然后就是记录路径，这里就是巧妙的地方；我还不太清除是怎么搞的，总之就是一共进行了k次操作，就相当于01全排列，首项是1，然后排到第一个符合的数的时候，这个数是第几个，它对应的二进制就是相应的串；这一题用bfs居然超时了；所以我打了个表，有一个不打表的做法；

代码如下：

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<queue>
 using namespace std;

 queue<int> q;
 int n;
 char a[][]={"","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","",""};
 /*
 int cou=0;
 void bfs()
 {
     while(!q.empty())
     {
         int t=q.front();
         q.pop();
             cou++;
         if(t%n==0)
             return;
         q.push(t*10%n);
         q.push((t*10+1)%n);
     }
 }
 int main()
 {
     freopen("out","w",stdout);
     for(int z=1;z<=200;z++)
     {
         n=z;
         while(!q.empty())
             q.pop();
         memset(a,0,sizeof(a));
         cou=0;
         a[0]=1;
        q.push(1);
        bfs();
        int i=0;
        while(cou)
        {
            a[i++]=cou%2;
            cou=cou/2;
        }
        printf("\"");
        for(int j=i-1;j>=0;j--)
        {
            printf("%d",a[j]);
        }
        printf("\",");
     }
     return 0;
 }*/

 int main()
 {
     while(cin>>n&&n)
     {
         printf("%s\n",a[n-]);
     }
     return ;
 }

 //Memory Time
 //2236K  32MS 

 #include<iostream>
 using namespace std;

 int mod[];  //保存每次mod n的余数
                   //由于198的余数序列是最长的
                   //经过反复二分验证，436905是能存储198余数序列的最少空间
                   //但POJ肯定又越界测试了...524286是AC的最低下限，不然铁定RE

 int main(int i)
 {
     int n;
     while(cin>>n)
     {
         if(!n)
             break;

         mod[]=%n;  //初始化，n倍数的最高位必是1

         for(i=;mod[i-]!=;i++)  //利用同余模定理，从前一步的余数mod[i/2]得到下一步的余数mod[i]
             mod[i]=(mod[i/]*+i%)%n;
                      //mod[i/2]*10+i%2模拟了BFS的双入口搜索
                      //当i为偶数时，+0，即取当前位数字为0  。为奇数时，则+1,即取当前位数字为1

         i--;
         int pm=;
         while(i)
         {
             mod[pm++]=i%;   //把*10操作转化为%2操作，逆向求倍数的每一位数字
             i/=;
         }
         while(pm)
             cout<<mod[--pm];  //倒序输出
         cout<<endl;
     }
     return ;
 }