Find The Multiple (poj 1426 bfs)

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18454		Accepted: 7461		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

Source

Dhaka 2002

题意：输入整数n，求一个只由1和0两个数字组成的十进制整数m，m是整数n（200以内）的k倍，且要求k最小。

思路：运用同余定理+bfs,看这位大神这一题的同余定理。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

struct Node
{
    int num;
    int mod;
    Node *pre;//记录路径
};

int N;
int visit[maxn];

void print(Node *x)//递归输出路径
{
    if (x->pre==NULL)
        printf("%d",x->num);
    else
    {
        print(x->pre);
        printf("%d",x->num);
    }
}

void bfs()
{
    int i;
    queue<Node>Q;
    memset(visit,0,sizeof(visit));
    Node sst,*now=new Node;
    visit[1]=1;
    while (!Q.empty())
        Q.pop();
    sst.mod=1;sst.num=1;sst.pre=NULL;
    Q.push(sst);
    while (!Q.empty())
    {
        Node *st=new Node;
        *st=Q.front();
        Q.pop();
        if (st->mod==0)
        {
            print(st);
            printf("\n");
            return ;
        }
        for (i=0;i<2;i++)
        {
            int xx=(st->mod*10+i)%N;
            if (!visit[xx])
            {
                now->mod=xx;
                now->num=i;
                now->pre=st;
                Q.push(*now);
                visit[xx]=1;
            }
        }
    }
    return ;
}

int main()
{
    while (scanf("%d",&N)&&N)
    {
        if (N==1)
        {
            printf("1\n");
            continue;
        }
        bfs();
    }
    return 0;
}