P2022 Interesting number

P2022 Interesting number topic description

Let's consider a set of positive integers from 1 to n. Let's arrange the elements in the collection in dictionary order, for example, when n=11, it should be: 1,10,11,2,3,4,5,6,7,8,9.

Defines the position of K in the N number as Q (n,k), for example q (11,2) = 4. Now given the integers K and M, it is required to find the smallest N, which makes Q (n,k) =m.

Input output Format input format:

The input file has only one row, which is two integers k and M.

Output format:

The output file has only one row, which is the smallest n, and outputs 0 if no such n exists.

Input and Output Sample input example # #:

Sample 1:2 4Sample 2:100,000,001 1000000000 here Sample 1 and 2 are separate two data points.

Sample # # of output:

Sample 1:11sample 2:1.,000,000,008,888,89e,+17

Description

"Data Contract"

40% of Data,1<=k,m<=10^5;

100% of the data, 1<=k,m<=10^9.

Feel:

I was thinking, was it violence? Double? Two points? ...... The result is the Tao number problem, I C ghost can make? See for yourself.

Exercises

Because the answer can be very large, the problem is obviously not enumerated, even with two points, time complexity {o[n (LOGN) ^2]} is also very large. We can set all the dictionary order than K small number of the first M-1 is X,n is equal to K and X maximum, how to find x? [Delete] is of course an enumeration. [/delete] We divide all dictionaries into infinitely large sets of smaller numbers than K. Any two-digit j,k in the set AI satisfies I=floor (log10 (j)) =floor (log10 (k)) (Floor (a) represents the integer portion of a, log10 (a) represents the base of 10, and A is true for a value), where the maximum value is AI. We can find that the left number I bit of K is pi,qi=∑pj* (i-j+1) (1<=j<=i), when J<=log10 (k), | aj|=qj-1,aj=qj-1; when J>log10 (K), | aj|=| A (j-1) |*10 (please prove it to the reader). This allows us to find the AI in the set of X, and x=ai+[m-∑| aj| (1<=j<=i)] -1. To find the time complexity of the set of X and to find the value of x after the set of X is O[LOG10 (n)], the total time complexity is O[LOG10 (n)].

AC Code:

#include <iostream>using namespacestd;Long LongK,m,i,number,n;intMain () {CIN>>k>>m;  for(i=1; i<=k;i*=Ten) number+=k/i-i+1; number--; if(number>=m| | K (i/Ten)==0&&number<m-1) cout<<0; if(number>=m| | K (i/Ten)==0&&number<m-1)return 0;  for(I=k-(i/Ten), n=k;number<m-1; i*=Ten, number+=i,n*=Ten); cout<<max (n-number+m-2, K); return 0;}

P2022 Interesting number