POJ 1426 Find The Multiple (广搜)

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20235 Accepted: 8203 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

题目大意：给一个数n，求最小的k，使得k是n的整数倍，且k的十进制表示只有0和1.

题目上说这样的k不会超过200位，确实是真的，不过这样太吓人了，其实在1~200里的全部n的结果k都是在64位之内的，它特意说不超过200位，一开始把我吓坏了，这题用广搜怎么可能做呢，太坑爹了这题。

广搜，其实这个全部数据里最大的也不过是n=198时，k=1111111111111111110（不用数了，一共18个1，1个0）

AC代码：

#include <stdio.h>#include <queue>using namespace std;typedef __int64 ll;void bfs(int n){queue<ll> que;que.push(1);while(!que.empty()){ll t=que.front();que.pop();if(t%n==0){printf("%I64d\n",t);return ;}que.push(t*10);que.push(t*10+1);}}int main(){int i,n;while(scanf("%d",&n),n)bfs(n);return 0;}

下面还有一个代码，写的基本上一致，只不过把bfs改成了__int64的带返回值，然后在main输出结果，W!A!了。。。。我不明白，路过的大神帮忙指点一下可好

#include <stdio.h>#include <queue>using namespace std;typedef __int64 ll;ll bfs(int n){queue<ll> que;que.push(1);while(!que.empty()){ll t=que.front();que.pop();if(t%n==0)return t;que.push(t*10);que.push(t*10+1);}}int main(){int i,n;while(scanf("%d",&n),n)printf("%I64d\n",bfs(n));return 0;}