POJ 1426 Find The Multiple【dfs】

Find The Multiple

Time Limit: 1000MS

Memory Limit: 10000K

Total Submissions: 22877

Accepted: 9431

Special Judge

Description

Given a positiveinteger n, write a program to find out a nonzero multiple m of n whose decimalrepresentation contains only the digits 0 and 1. You may assume that n is notgreater than 200 and there is a corresponding m containing no more than 100decimal digits.

Input

The input file maycontain 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 ofn in the input print a line containing the corresponding value of m. Thedecimal representation of m must not contain more than 100 digits. If there aremultiple solutions for a given value of n, any one of them is acceptable.

Sample Input

2

6

19

0

Sample Output

10

100100100100100100

111111111111111111

题意：

给定一个数N（1≤N≤200），求任意一个N的倍数M*N【这个是我设】，使得数的十进制位每位都是由0或1组成。任意输出一组解就行。题目保证有解。

思路：

__int64可以装下。那么最多到达20个十进制位，那么就可以直接dfs水过了=_=

实现代码：

 

#include <queue>#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)typedef long long LL;int N;bool suc;void dfs(LL M, int bit) {    if(bit >= 20 || suc) return;    if(M % N == 0) {        suc = true;        printf("%I64d\n", M);        return;    }    dfs(M * 10, bit + 1);    dfs(M * 10 + 1, bit + 1);}int main(){//    FIN;    while(~scanf("%d", &N) && N) {        suc = false;        dfs(1, 1);    }    return 0;}