POJ 1426 Find The Mutiple

Find The Mutiple
Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22958		Accepted: 9457		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

刚看到这题的时候被它的数据量给吓到了，总想着它DFS肯定会超时，而BFS肯定会超空间。后来发现这个题没有那么难，用一个初始值为1的数m。进行深搜或广搜（m*10；m*10+1），一步步进行操作就可以了。这个我是一次过的，至于网上有的博客说会超空间什么的我觉得不会，直接用long long 来存这个数就足够了，不需要开辟很大的数组进行广搜。至于深搜，对于步骤数要进行控制，即不能操作超过19次，否则会出错。嗯嗯，网上还有很好的剪枝方案，值得去看下。

DFS：
#include <iostream>
#include <cstdio>
using namespace std;
int flag,n;
void dfs(long long int m,int step)
{
if(flag||step==19)
return;
if(m%n==0)
{
printf("%lld\n",m);
flag=1;
return;
}
dfs(m*10,step+1);
dfs(m*10+1,step+1);
return;
}
int main()
{
while(scanf("%d",&n)&&n)
{
flag=0;
dfs(1,0);
}
return 0;
}
BFS：
#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
int n;
void bfs()
{
queue<long long int> Q;
long long int m=1;
Q.push(m);
while(!Q.empty())
{
m=Q.front();
Q.pop();
if(m%n==0)
{
printf("%lld\n",m);
return;
}
Q.push(m*10);
Q.push(m*10+1);
}
}
int main()
{
while(scanf("%d",&n)&&n)
{
bfs();
}
return 0;
}