POJ.Find The Multiple

题目

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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

大致题意

一个数N(N>0,N<200),找一个数是N的倍数，这个数是一个十进制数，它是由1,0，组成的，例如:100101010,就是一个这样的数。
思路：
很显然，这样的数不止有一个，因此，用深度优先搜索，找到了一个解即退出。用unsigned __int64 来储存其倍数x.不带符号的64位整数输出方式:
printf("%I64u\n",x);

代码：

#include <iostream>#include<cstring>#include<stack>#include<cstdio>#include<algorithm>#include<cmath>#include<queue>using namespace std;bool vis;int n;void dfs( unsigned __int64 x,int n,int dep){ if(vis)return;//这两个if一定不能颠倒。 //必须先检测是否已经输出x //只有检测并没有输出过后才能放心的输出 if(x%n==0) {    printf("%I64u\n",x);    vis=true;    return ; } //如果颠倒了之后 //如果有多个解的话会输出多个解 if(dep==19)//unsigned long long 的最大位数 return;    dfs(x*10,n,dep+1);    dfs(x*10+1,n,dep+1);}int main(){    while(cin>>n,n)    {        vis=false;        dfs(1,n,0);    }    return 0;}