POJ—1426

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32261 Accepted: 13470 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

【分析】

比较简单的一道搜索题...*10和*10+1两种情况，long long能吃的下最终情况..所以直接搜索就可以了

不过这里有一种，非常简洁的方式，因为只有两种情况*10或者*10+1，非常像二叉树，所以这里可以用哈夫曼思想，直接在数组中模拟搜索

【代码】

#include<iostream>  #include<stdio.h>  #include<string.h>  #include<queue>  using namespace std;    long long f[1000000];    int main()  {      int n;f[0]=0;    while(~scanf("%d",&n)&&n)      {          int i;          for(i=1;;i++)          {              f[i]=f[i/2]*10+i%2;              if(f[i]%n==0) break;          }          printf("%lld\n",f[i]);      }      return 0;  }  

dfs

#include <stdio.h>int n;long long ans=-1;void dfs(long long now,int deep){if (ans!=-1) return;if (deep>19) return;if (now%n==0){ans=now;return;}dfs(now*10,deep+1);if (ans!=-1) return;dfs(now*10+1,deep+1);}int main(){while (~scanf("%d",&n)&&n){ans=-1;dfs(1,1);printf("%lld\n",ans);}return 0;}