POJ_1426_FindTheMultiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21478 Accepted: 8797 Special Judge

Description

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

Source

Dhaka 2002

这个问题题目描述比较吓人，但事实上long long范围内是可以找到解的，

而且注意到题目要求输出any one of them is acceptable，所以。偷懒点用19层以内dfs就可以过了

//这问题真的是醉了，既然直接dfs可以的话……//不过感觉真的老老实实bfs貌似比较麻烦#include <iostream>#include <stdio.h>using namespace std;typedef unsigned long long LL;  //这里可以用long longint f;LL ans;void dfs(LL n,LL mu,int s){    if(f||(s==19))        return;    if(mu%n==0)    {        f=1;        ans=mu;        return;    }    else    {        dfs(n,mu*10,s+1);        dfs(n,mu*10+1,s+1);    }    return;}int main(){    int n;    while(1)    {        f=0;        scanf("%d",&n);        if(!n)            break;        LL nn=n;        dfs(nn,1,0);        printf("%I64u\n",ans);    }    return 0;}

bfs的方法以后学了数论相关的补充