Find The Multiple（数学+搜索）

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

26190

Sample Output

10100100100100100100
111111111111111111
思路1 dfs ：因为结果是n的倍数，所以必须有一个因子x，x*n = m；那么模拟乘法，留下最后一位，只不过最后一位只能是0或1，维护剩余的进位dat，当dat = 1 时，乘法就算完了。具体就是先对n处理一下，把后面的0去掉，并存起来。 for（i ：  0 - 9） ，如果n*i+dat个位是 1 或 0，满足题意， 那么dat/=10，并且记录一下个位的值。例如：n = 6，则第一步：dat = 0， dat+ n*5 末尾是0  更新 dat = 3保留 ans【k++】 = 0，第二步，dat= 3  dat+n*3末位是1，更新dat = 2 第三步回到上个例子；注意要剪枝，dat不可重复出现不然死循环。
#include <cstdio>#include <queue>#include <map>#include <climits>#include <cstring>#include <iostream>using namespace std;int ans[150];//记录最终结果map <int,int> a; // 标记dat是否有过.int n;bool ok;void dfs(int dat,int k){    if(ok || k > 50) return; // K不要取到100，可能会超内存，本来想分梯度，但是改了下50，竟然就过了    if(dat == 1)    {        ans[k] = dat;        for(int i = k; i >=0; i--)            printf("%d",ans[i]);        printf("\n");        ok = 1;        return;    }    for(int i = 0; i <= 9; i++)    {        if(((dat+n*i) % 10 ) / 2 == 0)        {            int tmp;            ans[k] = (dat+n*i) % 10;            tmp = (dat+n*i)/10;            if(a[tmp]) continue;            a[tmp] = 1;            dfs(tmp,k+1);            a[tmp] = 0;        }    }}int main(){    while(scanf("%d",&n)&&n)    {        a.clear();        a[0] = 1,ok=0;        int k = 0;        while((n%10) == 0)        {            ans[k++] = 0;            n/=10;        }        if(n == 1)        {            ans[k] = 1;            for(int i = k; i >=0; i--)                printf("%d",ans[i]);            printf("\n");            continue;        }        dfs(0,k);    }    return 0;}


后来见了个代码，直接bfs n 多花点时间应该也能过，试了试真过了，真是蒙了。
#include<cstdio>#include<vector>using namespace std;int find;void dfs(unsigned long long t,int n,int k){       if(find)return;       if(t%n==0)       {            printf("%I64u\n",t);            find=1;            return;       }       if(k==19)   return ;       dfs(t*10,n,k+1);       dfs(t*10+1,n,k+1);}int main(){    int n;    while(scanf("%d",&n),n)    {         find=0;         dfs(1,n,0);    }    return 0;}