POJ

题目链接：http://poj.org/problem?id=1426点击打开链接

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33048 Accepted: 13830 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

Source

Dhaka 2002

题意迷人

讲的是给你一个数 你要找一个十进制的数 只能由0和1组成

然后这个十进制的数要能整除这个给的数

分析过程：

1.先创立一个数为1 （因为第一个数不为0）将他入队

2.每次从队中取出首元素进行判断 然后就是*10与（*10+1）的两个方向的广搜

3.然后计算 当前这个数两个方向的值 与给出数的余数 如果不为0 就看这个余数是否被标记过 如果标记过就跳过 没有就标记 然后入队 重复2 直到队列为空或发现余数为0

这样的标记化需要注意要记录每个操作 用vector的赋值可以快速完成 

这里将1记为*10

将2记为+1；（因此（*10+1）1和2的操作都有记录）

最后根据记录的操作次数进行计算即可

此时应该也是保证所计算的数为最小的

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits.h>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;int main(){    int n=0;    while(~scanf("%d",&n)&&n)    {        int book[n+n];        for(int i=0;i<=n+n;i++)            book[i]=0;        vector<int > s[n+1];        vector<int > ss;        int mid=1;        queue <int > q;        while(!q.empty())            q.pop();        for(int i=0;i<n+1;i++)        s[i].clear();        ss.clear();        q.push(1);        while(!q.empty())        {            int mid=q.front();                q.pop();            ss=s[mid];            if(mid==0)            {                break;            }            if(!book[mid*10%n])            {                book[mid*10%n]=1;                s[mid*10%n]=ss;                s[mid*10%n].push_back(1);                q.push(mid*10%n);            }            if(!book[(mid*10+1)%n])            {                book[(mid*10+1)%n]=1;                s[(mid*10+1)%n]=ss;                s[(mid*10+1)%n].push_back(1);                s[(mid*10+1)%n].push_back(2);                q.push((mid*10+1)%n);            }        }        long long int ans=1;        for(int i=0;i<ss.size();i++)        {            if(ss[i]==1)                ans*=10;            else                ans+=1;        }        cout << ans<< endl;    }}