poj 1426 Find The Multiple

標籤：

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

題意：給出一個整數n，(1 <= n <= 200)。求出任意一個它的倍數m，要求m必須只由十進位的‘0‘或‘1‘組成。

     如果搜到m則輸出，否則搜尋m×10和m×10+1直到得出答案

 1 #include <iostream> 2 #include <stack> 3 #include <queue> 4 #include <cstdio> 5 using namespace std; 6 #define LL unsigned long long 7 int n; 8 bool flag; 9 void dfs(LL x,int step)10 {11     if(flag||step==19)//搜尋到或者到第19步時返回,因為第20層就超出了unsigned long long範圍12         return ;13     if(x%n==0)14     {  //發現輸出答案，並標記15         printf("%llu\n",x);16         flag=true;17         return ;18     }19     dfs(x*10,step+1);20     dfs(x*10+1,step+1);21     return ;22 }23 int main()24 {25     while(scanf("%d",&n),n)26     {27         flag=false; //標記是否找到題意之中的m28         dfs(1,0); // 從1開始搜尋n的倍數29     }30     return 0;31 }

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poj 1426 Find The Multiple