Find The Multiple（dfs）

Find The Multiple Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 28409 Accepted: 11784 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

2 6 19 0

Sample Output

10 100100100100100100 111111111111111111

Source

题目大意是给出一个数n，找出一个数要求是n的倍数，并且这个数的十进制只由1和0组成，明显这样的数不止一个（如果，满足条件一定会有m×10也满足，故不止一种），题目要求输出任意一个满足该条件的m 对于数据1，可知2×5=10，故答案可以得出是10（当然，100，1000...也满足，但是special judge，只用输出一个满足条件的解），其他数据也同理。 #include <iostream> #include <cstdio> using namespace std; int n,flag; void dfs(unsigned long long x,int k) { if(flag||k==19) { return ; } if(x%n==0) { printf("%I64u\n",x); flag=1; return; } dfs(x*10,k+1); dfs(x*10+1,k+1); } int main() { while(~scanf("%d",&n)&&n) { flag=0; dfs(1,0); } return 0; } 因为n最大200，以为最后的结果会非常大，无符号整数可以存下结果，如果搜到m则输出，否则搜索x×10和x×10+1直到得出答案//只有1和0