POJ 1426 Find The Mutiple

Find The Mutiple

 Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal 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;}