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%7==0) return true;
    while(n) {
        if(n%10==7) return true;
        n/=10;
    }
    return false;
}
long long a[100000010];
long long  b[100010000];
int main(){
    int i,c=0,cc=0;
    //freopen("in.txt","r",stdin);
    for(i=1;i<=100000000;i++)
      if(fun(i)) a[++c]=i;
    for(i=1;i<c;i++)
       if(!fun(i))
         { b[++cc]=a[i];}
    long long  n;
    while(cin>>n){
        if(n>cc-1) cout<<"??";
        else 
        cout<<b[n]<<endl;}
    return 0;
}

官方标程：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ULL unsigned long long
const ULL Maxn=((1uLL<<64uLL)-1);
ULL dp[22][10][2];
int digit[24];
ULL dfs(int pos,int pre,int flag,bool limit){
    if(pos==-1) return flag||pre==0;
    if(!limit&&dp[pos][pre][flag] != -1) return dp[pos][pre][flag];
    int end = limit?digit[pos]:9;
    int fflag,ppre;
    ULL ans=0;
    for(int i = 0;i <= end;i++){
        fflag = (i==7)||flag;
        ppre=(pre*10+i)%7;
        ans+=dfs(pos-1,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 = 0;
    while(n){
        digit[pos++] = n%10;
        n /= 10;
        if(!n) break;
    }
    return dfs(pos-1,0,0,1)-1;
}
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,-1,sizeof(dp));
    ULL n;
    while(cin>>n){
            ULL l = 0, r= Maxn,mid;
            while(l <= r){
                mid = ((r-l)>>1)+l;
                if(find(mid)<n) l = mid+1;
                else r = mid-1;
            }
            ULL ans = r+1;
            cout<<ans<<endl;
    }
    return 0;
}