POJ-1426-Find The Multiple (BFS +DFS)两种解法

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 35878 Accepted: 14986 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

题解：还是一道简单的dfs题，已知n 求m 是n的倍数 ，m（十进制中只有包含0和1 ） 不会超过100个数位。unsigned __int64 可存放结果，包含两条搜索路径，x*10 和 x*10+1 这两条.

当然也可以用BFS来做,见AC2.

AC1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool vis;
int n;
void dfs(unsigned __int64 x,int n,int k)
{
if(vis) return;
if(x%n==0)
{
printf("%I64u\n",x);
vis=true;
return;
}
if(k==19) return;
dfs(x*10,n,k+1);
dfs(x*10+1,n,k+1);
}
int main()
{

while(scanf("%d",&n))
{

if(n==0) break;
vis=false;
dfs(1,n,0);
}
return 0;
}

AC2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
int n;
void bfs()
{

queue<long long > q;
while(!q.empty())
q.pop();
q.push(1);
while(1)
{
long long tmp=q.front();
if(tmp%n==0)
{
cout<<tmp<<endl;
return;
}

q.pop();
q.push(tmp*10);
q.push(tmp*10+1);
}
}
int main()
{

while(scanf("%d",&n),n)
bfs();
return 0;

}