The fourth session of Shandong Province A-number and B-number [digital dp+ Two-point answer] dynamic planning

Topic Link: Hrbust's OJ on the topic

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A-number and B-number
Time limit:1000 MS Memory limit:32768 K
Total submit:22 (8 users) Total Accepted:5 (5 users) rating:special Judge:no
Description
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 an integer whose decimal form contains 7 or it can be divided by 7. We can write down the first ten 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 and A[k] (that's K is A a-number.) Like, the 7th A-number and 7 are also an a-number so the (A[7]) are not A b-number. We also can write down the first ten 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 a integer N, please output the Nth b-number.

Input
The input consists of multiple test cases.

For each test case, there would be a positive integer N as the description. (0<n<263−1) (0

Output
For each test case, the output of an integer indicating the Nth b-number.

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

Sample Input
1
7
100
Sample Output
7
37
470

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Main topic:
A array is a series of folk games "knock Seven"
B array is {x∣∣xi∈{a} and I∉{a}} \{x \big| x_i \in \{a\} and I\notin \{a\}\}
Then output the nth element in the B array: Bn b_n

Problem Solving Ideas:

If I were to find the first element I would not be able to ask whether it was an array of a or B (I would not)

But he's got the I-I element. We can think of it in a different way.

First we can use an entry-level digital DP to count the number of elements less than n N,

Then the smallest n of the x element in the [1,n] is the first X element

Then we can use the two-point answer to count.

For a arrays we're good at digital DP
But we can't just ask for a B-array.

However, considering the relationship between A and B arrays, it is found that as long as the results of CAL (X) Statistics ans is statistically subtracted;

That

A = cal (x);
B = a-cal (A);

Attention
Two-point overflow problem
Maximum Value (1<<63)-1,

Attached code
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#include <bits/stdc++.h> using namespace std;
typedef long long int LL;
/********************************/int num[20];

LL dp[20][8][2]; LL dfs (int pos,int mod,int limit,int status) {if (pos<0) return (status| |
    mod%7==0); if (Dp[pos][mod][status]!=-1&&!limit) return dp[pos][mod][status];
    Forget to write limit Meng forced 1000 days there is no int endi=9;

    if (limit) endi=num[pos];
    LL res = 0; for (int i=0;i<=endi;i++) Res+=dfs (pos-1, (mod*10+i)%7,limit&&i==endi,i==7| |

    status);
    if (!limit) dp[pos][mod][status] = res;
return res;
    } ll Cal (ll x) {if (!x) return 0;
    int len=0;
    while (x) num[len++]=x%10,x/=10;
Return DFS (len-1,0,1,0)-1;  } void Solve (ll x) {ll l=7,r= (1ll<<63) -1,mid,ans=-1;
        R written in 1e18 WA Meng has no,,,, while (l<=r) {mid= ((r-l) >>1) +l;
        LL tmp = CAL (mid);
        TMP-= CAL (TMP);
        if (tmp>=x) ans=mid,r=mid-1;
    else l=mid+1;
} printf ("%lld\n", ans); } int main () {///printf ("%lld\n ", ~ (1ll<<63));
    Memset (Dp,-1,sizeof (DP));
    LL N;
    while (~SCANF ("%lld", &n)) solve (n);
return 0; }