HDU 1664 Different Digits

Different Digits

Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 464    Accepted Submission(s): 106

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

 

Source

2004 Asia Regional Shanghai

 

Recommend

xhd

 

思路：根据抽屉原理，可以先搜只有一位数字的，然后从小到大搜索两位数字的

但是我在搜索两位数字的时候，采用暴力搜索，结果毫无疑问的超时了，悲剧

了，在3170上卡了很久，而且在以30,40,70,80,20,90，结尾时时间开销有点

大

 

代码：

超时（暴搜）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int n;
int flag,temp,sum[100],step;
int num[100];
void BFS(int a)
{
    int biji;
    int map[10000];
    int qmap = 0,p;
    while(a > 9)
    {
        p = a % 10;
        a = a / 10;
        map[qmap ++] = p;
    }
    map[qmap] = a;
    int mmm[10000];
    memset(mmm,0,sizeof(mmm));
    for(int q = 0;q <= qmap;q ++)
        mmm[q] = map[q];
    int shumu = 0;
    for(int k = 0;k <= 9;k ++)
                num[k] = k;
    for(int l = 0;l <= qmap;l ++)
    {
        if(num[map[l]] != -1)
        {
                shumu ++;
                num[map[l]] = -1;
        }
    }
    if(shumu == 2)
    {
        for(int m = qmap;m >= 0;m --)
                printf("%d",map[m]);
        printf(" ");
        return ;
    }
    while(1)
    {
        biji = 0;
        for(int j = 0;j <= qmap;j ++)
        {
               map[j] = map[j] + mmm[j] + biji;
               biji = map[j] / 10;
               map[j] = map[j] % 10;    
        }
        while(biji > 10)
        {
           qmap ++;
           map[qmap] = biji % 10;
           biji = biji / 10;
        }
        if(biji)
        {
            qmap ++;
            map[qmap] = biji;
        }
        int shumu = 0;
        for(int k = 0;k <= 9;k ++)
                    num[k] = k;
        for(int l = 0;l <= qmap;l ++)
        {
            if(num[map[l]] != -1)
            {
                shumu ++;
                num[map[l]] = -1;
            }
        }
        if(shumu == 2)
        {
            for(int m = qmap;m >= 0;m --)
                   printf("%d",map[m]);
            printf(" ");
            return ;
        }
    }
}
int main()
{
    //while(scanf("%d",&n),n != 0)
    for(int n = 3160;n <= 3180;n ++)
    {
        printf("%d ",n);
        flag = 0;
        temp = 0;
        memset(sum,0,sizeof(sum));
        for(int i = 1;i <= 100000;i ++)
        {
            for(int j = 1;j <= 9;j ++)
            {
                temp = j;
                sum[j] = sum[j] * 10 + j;
                sum[j] = sum[j] % n;
                if(sum[j] == 0 || sum[j] % n == 0)
                {
                    flag = 1;
                    step = i;
                    break;
                }
            }
            if(flag)
              break;
        }
        if(flag)
        {
            for(int k = 1;k <= step;k ++)
               printf("%d",temp);
            printf(" ");
        }
        else
        {
            BFS(n);
        }
    }
    return 0;
}