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Find The Multiple
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 Source |
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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;
const int INF = 1000010;
using namespace std;
int n;
struct node
{
string s;
int mod;//保留前一个的摸值
};
bool vis[222];
queue<node>que;
void bfs()
{
while(que.size())
que.pop();
memset(vis,0,sizeof vis);
node a,t;
a.mod=1;
a.s="1";
que.push(a);
vis[1]=1;
while(que.size())
{
a=que.front();
que.pop();
if(a.mod==0)
{
cout<<a.s<<endl;
return;
}
for(int i=0; i<2; i++)
{
t.mod=a.mod*10+i;
t.mod%=n;
if(!vis[t.mod])
{
if(i)
t.s=a.s+"1";
else
t.s=a.s+"0";
que.push(t);
vis[t.mod]=1;
}
}
}
}
int main()
{
while(cin>>n&&n)
{
bfs();
}
return 0;
}
poj 1426 Find The Multiple(bfs)
原文地址：http://blog.csdn.net/acm_baihuzi/article/details/42346147
