hdu 1664 Different Digits

Different Digits

Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1138    Accepted Submission(s): 296

Problem Description

Given a positive integer n, your task is to find a positive integer m, which is a multiple of n, and that m contains the least number of different digits when represented in decimal. For example, number 1334 contains three different digits 1, 3 and 4.

 

Input

The input consists of no more than 50 test cases. Each test case has only one line, which contains a positive integer n ( 1<=n < 65536). There are no blank lines between cases. A line with a single `0' terminates the input.

 

Output

For each test case, you should output one line, which contains m. If there are several possible results, you should output the smallest one. Do not output blank lines between cases.

 

Sample Input

   
   
   
   

    
    
    
    7 
15 
16 
101 
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7
555
16
1111
   
   
   
   

题意：告诉一个数n,求一个他的倍数m，使得m由最少的不同数字组成。

思路：对于任意一个n，都可以找到一个他的倍数，并且这个倍数只有最多2种不同的数字组成。所以，可以分成1种和2种讨论下。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

struct Node
{
    int mod;
    string ans;
}f,t;

int n;
string ans,tmp;
int num[2];
int vis[70000];

int bfs1()
{
    int flag=0;
    for(int i=1;i<=9;i++)
    {
        f.mod=i%n;
        f.ans="";
        f.ans+=i+'0';

        while(f.ans.size()<n && f.mod!=0)
        {
            f.ans+=i+'0';
            f.mod=(f.mod*10+i)%n;
        }
        if(f.mod==0)
        {
            if(flag==0) {tmp=f.ans;flag=1;}
            else if(f.ans.size()<tmp.size()) tmp=f.ans;
            else if(f.ans.size()==tmp.size() && f.ans<tmp) tmp=f.ans;
        }
    }
    if(flag) {ans=tmp;return 1;}
    else return 0;
}

void bfs2()
{
    memset(vis,0,sizeof vis);
    queue<Node>q;

    for(int i=0;i<2;i++)
    {
        if(num[i]==0) continue;
        f.mod=num[i]%n;
        f.ans="";
        f.ans+=num[i]+'0';
        vis[f.mod]=1;
        q.push(f);
    }

    int flag=0;

    while(!q.empty())
    {
        f=q.front();
        q.pop();

        if(flag)
        {
            if(f.mod==0)
            {
                if(tmp.size()>f.ans.size()) {tmp=f.ans;continue;}
                else if(tmp.size()==f.ans.size() && tmp>f.ans) {tmp=f.ans;continue;}
            }

        }
        else if(f.mod==0)
        {
            flag=1;
            tmp=f.ans;
            continue;
        }

        for(int i=0;i<2;i++)
        {
            t=f;
            t.ans+=num[i]+'0';
            t.mod=(f.mod*10+num[i])%n;
            if(vis[t.mod]) continue;
            vis[t.mod]=1;
            q.push(t);
        }

    }

    if(ans=="") ans=tmp;
    else if(ans.size()>tmp.size()) ans=tmp;
    else if(ans.size()==tmp.size() && ans>tmp) ans=tmp;

}

int main()
{
    while(scanf("%d",&n),n)
    {
        ans="";

        if(bfs1());
        else
        {
            for(int i=0;i<9;i++)
                for(int j=i+1;j<=9;j++)
                {
                    num[0]=i;num[1]=j;
                    bfs2();
                }
        }
        cout<<ans<<endl;
    }
    return 0;
}