poj1426 Find The Multiple

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

10100100100100100100111111111111111111

深度搜索

#include<cstdio>#include<cstring>using namespace std;typedef unsigned long long ll;bool vis;void dfs(ll x,int n,int k){    if(vis)        return ;    if(x%n==0)    {        vis=true;        printf("%llu\n",x);        return ;    }    if(k==19)        return ;    dfs(x*10,n,k+1);    dfs(x*10+1,n,k+1);}int main(){    int n;    while(scanf("%d",&n),n)    {        vis=false;        dfs(1,n,0);    }    return 0;}

 广搜    我用c++交一直超内存

 换成g++ 过了（心累 ） 

#include<cstdio>#include<queue>using namespace std;typedef long long ll;queue<ll>que;void bfs(int n){        while(!que.empty())            que.pop();        que.push(1);        while(1)        {            ll tmp=que.front();            if(tmp%n==0)            {                printf("%lld\n",tmp);                break;            }            que.pop();            que.push(tmp*10);            que.push(tmp*10+1);        }}int main(){    int n;    while(scanf("%d",&n),n)    {        bfs(n);    }    return 0;}