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：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 bool fun(int n){
6 if(n%7==0) return true;
7 while(n) {
8 if(n%10==7) return true;
9 n/=10;
10 }
11 return false;
12 }
13 long long a[100000010];
14 long long b[100010000];
15 int main(){
16 int i,c=0,cc=0;
17 //freopen("in.txt","r",stdin);
18 for(i=1;i<=100000000;i++)
19 if(fun(i)) a[++c]=i;
20 for(i=1;i<c;i++)
21 if(!fun(i))
22 { b[++cc]=a[i];}
23 long long n;
24 while(cin>>n){
25 if(n>cc-1) cout<<"??";
26 else
27 cout<<b[n]<<endl;}
28 return 0;
29 }
官方标程：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 #define ULL unsigned long long
6 const ULL Maxn=((1uLL<<64uLL)-1);
7 ULL dp[22][10][2];
8 int digit[24];
9 ULL dfs(int pos,int pre,int flag,bool limit){
10 if(pos==-1) return flag||pre==0;
11 if(!limit&&dp[pos][pre][flag] != -1) return dp[pos][pre][flag];
12 int end = limit?digit[pos]:9;
13 int fflag,ppre;
14 ULL ans=0;
15 for(int i = 0;i <= end;i++){
16 fflag = (i==7)||flag;
17 ppre=(pre*10+i)%7;
18 ans+=dfs(pos-1,ppre,fflag,limit&&i==end);
19 }
20 if(!limit) dp[pos][pre][flag]=ans;
21 return ans;
22 }
23 ULL solve(ULL n){//找到n以下有几个A-number
24 int pos = 0;
25 while(n){
26 digit[pos++] = n%10;
27 n /= 10;
28 if(!n) break;
29 }
30 return dfs(pos-1,0,0,1)-1;
31 }
32 ULL find(ULL n){//找到n以下有几个B-number
33 ULL t = solve(n);//表示n包括n以下在A中有t个
34 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
35 return tt;
36 }
37 int main()
38 {
39 memset(dp,-1,sizeof(dp));
40 ULL n;
41 while(cin>>n){
42 ULL l = 0, r= Maxn,mid;
43 while(l <= r){
44 mid = ((r-l)>>1)+l;
45 if(find(mid)<n) l = mid+1;
46 else r = mid-1;
47 }
48 ULL ans = r+1;
49 cout<<ans<<endl;
50 }
51 return 0;
52 }