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

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

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

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

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==){
             cout<<b<<endl;
             while(!s.empty()){
                 s.pop();
             }
             return ;
         }
         else{
             s.push(b*);
             s.push(b*+);
         }
     }
     return ;
 }

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