sdutoj 2609 A-Number and B-Number

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2609

A-Number and B-Number

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

Tom is very interested in number problem. Nowadays he is thinking of a problem about A-number and B-number.
    A-number is a positive integer whose decimal form contains 7 or it can be divided by 7. We can write down the first 10 A-number ( a[i] is the ith A-number) 
         {a[1]=7,a[2]=14,a[3]=17,a[4]=21,a[5]=27,a[6]=28,a[7]=35,a[8]=37,a[9]=42,a[10]=47};
    B-number is Sub-sequence of A-number which contains all A-number but a[k] ( that k is a  A-number.)  Like 35, is the 7th A-number and 7 is also an A-number so the 35 ( a[7] ) is not a B-number. We also can write down the first 10 B-number.

{b[1]=7,b[2]=14,b[3]=17,b[4]=21,b[5]=27,b[6]=28,b[7]=37,b[8]=42,b[9]=47,b[10]=49};
    Now Given an integer N, please output the Nth B-number

输入

The input consists of multiple test cases.

For each test case, there will be a positive integer N as the description.

输出

For each test case, output an integer indicating the Nth B-number.

You can assume the result will be no more then 2^63-1.

示例输入

1
7
100

示例输出

7
37
470

提示

 

来源

 2013年山东省第四届ACM大学生程序设计竞赛

示例程序

分析：

一开始想到打表，结果超时咯，后来看了标程用了搜索写的，，，orz

超时代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 bool fun(int n){
     if(n%==) return true;
     while(n) {
         if(n%==) return true;
         n/=;
     }
     return false;
 }
 long long a[];
 long long  b[];
 int main(){
     int i,c=,cc=;
     //freopen("in.txt","r",stdin);
     for(i=;i<=;i++)
       if(fun(i)) a[++c]=i;
     for(i=;i<c;i++)
        if(!fun(i))
          { b[++cc]=a[i];}
     long long  n;
     while(cin>>n){
         if(n>cc-) cout<<"??";
         else
         cout<<b[n]<<endl;}
     return ;
 }

官方标程：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define ULL unsigned long long
 const ULL Maxn=((1uLL<<64uLL)-);
 ULL dp[][][];
 int digit[];
 ULL dfs(int pos,int pre,int flag,bool limit){
     if(pos==-) return flag||pre==;
     if(!limit&&dp[pos][pre][flag] != -) return dp[pos][pre][flag];
     int end = limit?digit[pos]:;
     int fflag,ppre;
     ULL ans=;
     for(int i = ;i <= end;i++){
         fflag = (i==)||flag;
         ppre=(pre*+i)%;
         ans+=dfs(pos-,ppre,fflag,limit&&i==end);
     }
     if(!limit) dp[pos][pre][flag]=ans;
     return ans;
 }
 ULL solve(ULL n){//找到n以下有几个A-number
     int pos = ;
     while(n){
         digit[pos++] = n%;
         n /= ;
         if(!n) break;
     }
     return dfs(pos-,,,)-;
 }
 ULL find(ULL n){//找到n以下有几个B-number
     ULL t = solve(n);//表示n包括n以下在A中有t个
     ULL tt = t - solve(t);//表示t(包括第t个)个A_number中有几个是符合在B中的，一直找到tt刚好等于n为止
     return tt;
 }
 int main()
 {
     memset(dp,-,sizeof(dp));
     ULL n;
     while(cin>>n){
             ULL l = , r= Maxn,mid;
             while(l <= r){
                 mid = ((r-l)>>)+l;
                 if(find(mid)<n) l = mid+;
                 else r = mid-;
             }
             ULL ans = r+;
             cout<<ans<<endl;
     }
     return ;
 }