hdu1426搜索

http://poj.org/problem?id=1426

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

题意，找出任意一个由0和1组成的数能整除n。

一开始bfs超时

#include <iostream>
#include <queue>
using namespace std;
int n;
long long num;
long long bfs()
{
queue<long long>q;
q.push(1);
while(!q.empty())
{
num=q.front();
q.pop();
if(num%n==0)
break;
q.push(10*num);
q.push(10*num+1);
}
return num;
}
int main()
{while(cin>>n&&n)
{
bfs();
cout<<num<<endl;
}
return 0;
}

dfsAC

#include <iostream>
#include <cstdio>
using namespace std;

bool found;
int n;

void dfs(unsigned long long num,int deep)
{
if(found)
return;
if(num%n==0)
{
found=true;
cout<<num<<endl;
return;
}
if(deep>=19)//搜索第19就返回.20位装不下了
return;
dfs(num*10,deep+1);
dfs(num*10+1,deep+1);
}

int main()
{
while(cin>>n&&n)
{
found=false;
dfs(1,0);
}
return 0;
}

生找

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;

int n;
long long q[maxn];

long long work()
{
int front = 0;
int rear = 0;

q[rear++] = 1;
while (1)
{
long long temp = q[front++];
if ((temp * 10) % n == 0)
return temp * 10;
if ((temp * 10 + 1) % n == 0)
return temp * 10 + 1;
q[rear++] = temp * 10;
q[rear++] = temp * 10 + 1;
}
return -1;
}

int main()
{
//    for (n = 1; n <= 200; n++)
//        printf("%lld\n", work());
while (scanf("%d", &n), n)
printf("%lld\n", work());
}