[Poj 1426] Find The Multiple BFS

Find The Multiple
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 24596 Accepted: 10128 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’

用BFS来搜索满足条件的数，注意必须使用long long

#include<iostream>#include<stdio.h>#include<queue>using namespace std;int x;long long ans;int flag;struct node{    long long y;    int step;    node(long long yy,int st)    {        y=yy;        step=st;    }node(){}};queue<node> q;void bfs(){    while(!q.empty()) q.pop();    q.push(node(0,1));    q.push(node(1,1));    if(x==1) flag=1;    if(x==1) ans=1;    while(!q.empty())    {        if(flag) break;        node now=q.front();        //cout<<now.y<<endl;        q.pop();        for(int k=0;k<=1;k++)        {        //cout<<now.step<<endl;            long long mx=0;            if(k==1) mx++;            if(k==1)            for(int i=1;i<=now.step;i++)   mx*=(long long)10;            //cout<<"   "<<mx<<endl;            mx+=now.y;            if(mx!=0&&mx%x==0)            {                ans=mx;                flag=1;            }            q.push(node(mx,now.step+1));        }       }   }int main(){    while(scanf("%d",&x))    {        flag=0;        if(x==0) return 0;        bfs();        printf("%lld\n",ans);    }}