hdu 4294 Multiple BFS

最新推荐文章于 2019-03-22 20:07:23 发布

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最新推荐文章于 2019-03-22 20:07:23 发布

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分类专栏：搜索-DFS|BFS

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本文介绍了一道算法题的解决方法，该题要求找到一个正整数N的最小整数倍M，使得M在K进制表示下包含最少的不同数字，并详细解释了解题思路及实现代码。

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Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 954 Accepted Submission(s): 230

Problem Description

　　Given a positive integer N, you’re to solve the following problem:
　　Find a positive multiple of N, says M, that contains minimal number of different digits in base-K notation. If there’re several solutions, you should output the numerical smallest one. By saying numerical smallest one, we compar their numerical value, so 0xA _hex < 11 _dec.
　　You may assume that 1 <= N <= 10 ⁴ and 2 <= K <= 10.

Input

　　There’re several (less than 50) test cases, one case per line.
　　For each test case, there is a line with two integers separated by a single space, N and K.
　　Please process until EOF (End Of File).

Output

　　For each test case, you should print a single integer one line, representing M in base-K notation,the answer.

Sample Input

Sample Output

Source

2012 ACM/ICPC Asia Regional Chengdu Online

题目大意：求N的最小整数倍在k进制下不同的数字的个数是最小的那个倍数

解题思路：对于a,aa,aaa,....,aaa..aa这n个数字的情况，如果有一个mod n为0，那么最小的数字个数为1，否则，在这n个数字中，必然存在2个数mod n是同余的，所以必然存在2个的差值aaaa000这种形式的数mod n为0，所以答案最多只会有两个数字。

对于一个数字的情况，直接暴力即可，对于两个的情况，枚举2个数字，用bfs模拟2个数字的排列，注意要以和mod n的值作为状态，这同一个mod n的值对应一个值最小的答案。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <queue>
#include <string>
#include <algorithm>
using namespace std;
int n,k;
struct node{
    int mod,w,x;
    node(){}
    node(int mod,int w,int x):mod(mod),w(w),x(x){}
}now;
string ret;
int pre[10010];
int s[10010];
bool vis[10010];
void bfs(){
    for(int i=1;i<k;i++){
        for(int j=0;j<i;j++){
            int ans=-1;
            queue<node> q;
            int z=0;
            memset(vis,0,sizeof vis);
            q.push(node(i,1,z));
            pre[0]=-1,s[0]=i,z++;
            vis[i]=1;
            if(j) q.push(node(j,1,z));
            pre[1]=-1,s[1]=j,z++;
            while(!q.empty()){
                now=q.front();
                q.pop();
                if(now.mod==0){
                    ans=now.x;
                    break;
                }
                if(!vis[(now.mod*k+j)%n]){
                    vis[(now.mod*k+j)%n]=1;
                    pre[z]=now.x;
                    s[z]=j;
                    q.push(node((now.mod*k+j)%n,now.w+1,z));
                    z++;
                }
                if(!vis[(now.mod*k+i)%n]){
                    vis[(now.mod*k+i)%n]=1;
                    pre[z]=now.x;
                    s[z]=i;
                    q.push(node((now.mod*k+i)%n,now.w+1,z));
                    z++;
                }
            }
            if(ans==-1) continue;
            string ss="";
            for(int i=ans;i!=-1;i=pre[i]){
                ss.push_back(s[i]+'0');
            }
            int l=ss.size();
            for(int i=0;i<l/2;i++) swap(ss[i],ss[l-1-i]);
            if(ss.size()<ret.size()){
                ret=ss;
            }else if(ss.size()==ret.size()){
                if(ss<ret){
                    ret=ss;
                }
            }
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&k))
    {
        ret="";
        for(int i=0;i<=n;i++) ret.push_back('9');
        string tmp;
        int mod;
        bool flag=0;
        for(int i=1;i<k;i++){
            tmp="";
            mod=0;
            for(int j=0;j<n;j++){
                tmp.push_back('0'+i);
                mod=(mod*k+i)%n;
                if(mod==0){
                    if(tmp.size()<ret.size()){
                        ret=tmp;
                        flag=1;
                    }else if(tmp.size()==ret.size()){
                        if(tmp<ret) ret=tmp,flag=1;
                    }
                    break;
                }
            }
        }
        if(!flag) bfs();
        printf("%s\n",ret.c_str());
    }
    return 0;
}
</span>