Find The Multiple（bfs）

Find The Multiple

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 43   Accepted Submission(s) : 21

Special Judge

Problem 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

PKU

题意：构造一个十进制由0和1组成的整数m，让m能够被n整除；题目中给出了m是小于等于100位的。

思路：bfs可以做出的，只是100位这个条件让我很头疼，我先用string试了试交了之后是超时，后来看了看答案用long long也给过了，所以m肯定没有100位的；

string类型代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

queue<string>s;
int a,c;
string b;


int chu(string s)
{
    c=0;
    for(int i=0;i<s.length();i++){
        c*=10;
        c+=s[i]-'0';
        c%=a;
    }
    if(c==0)
        return 1;
    else
        return 0;
}


int bfs()
{
    while(!s.empty()){
        b=s.front();
        s.pop();
        if(b.length()>100)
            continue;
        if(chu(b+'1')==1){
            cout<<b+'1'<<endl;
            while(!s.empty()){
                s.pop();
            }
            return 0;
        }
        else{
            s.push(b+'1');
        }

        if(chu(b+'0')==1){
            cout<<b+'0'<<endl;
            while(!s.empty()){
                s.pop();
            }
            return 0;
        }
        else{
            s.push(b+'0');
        }
    }
    return 0;
}


int main()
{
//    freopen("input.txt","r",stdin);
    while(cin>>a){
        if(a==0)
            break;
        else{
            s.push("1");
            bfs();
        }
    }
    return 0;
}

long long类型代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

queue<long long>s;
int a,c;
long long b;



int bfs()
{
    while(!s.empty()){
        b=s.front();
        s.pop();
        if(b%a==0){
            cout<<b<<endl;
            while(!s.empty()){
                s.pop();
            }
            return 0;
        }
        else{
            s.push(b*10);
            s.push(b*10+1);
        }
    }
    return 0;
}


int main()
{
//    freopen("input.txt","r",stdin);
    while(cin>>a){
        if(a==0)
            break;
        else{
            s.push(1);
            bfs();
        }
    }
    return 0;
}