Ultraviolet A 10706-Number Sequence

There is an array s [1] = 1, s [2] = 1 2 ,....... s [k] = 1 .... k. a given n is required to represent the nth digit of the array and the nth digit is required. For example, when n is 1, when 1 n is 2, it is 1, and when n is 3, it is 2.

Solution: 1: Train of Thought: prepare tables and search
2: The data given by the question can discover some patterns
S [1]: 1 Total number of digits 1
S [2]: 1 2 Total digits 1 + 2 = 3
S [3]: 1 2 3 Total digits 1 + 2 + 3 = 6
.
.
.
S [9]: 1 2 3 4 5 6 7 8 9 total digits 1 + 2 + 3 +... + 9 = 45
S [10]: 1 2 3 4 5 6 7 8 9 10 Total digits 45 + 1 + 2 + 3 +... + 9 + 1 + 0 = 56
S [K]: 1 2 3 4... k total number of digits num [k-1] + 1 + 2 + ......
Use num [k] To save the total number of digits when the value is k.
Therefore, if we can obtain all the data in advance to form a table, then we can directly search for the data in the input position, and then find the position. Due to the increasing trend above, I can infer that when k to 100000, it will exceed int, So you can open such a large array.
3: During the table creation process, I used to enumerate the number of digits of the current value. For example, if the number of digits is 1, there are 9 digits in total from 1 to 9, if the number of digits is 2, there will be 10-99 Public 90 ....... assume that the current value is n, then num [n] is obtained based on the above rule, and so on.
4: search, O (n) time complexity, as long as find num [I-1] <n, num [I]> = n, then we can know that this number exists in s [n], subtract n minus num [I-1], we can know the number of the required number in s [n, then let's judge them one by one. Note that n may be multiple digits such as 1234, so you need to split it first, that is, the maximum bit.
5. Note: Because the n value of this question is 2147483647 at most, we need to open the num array as long, and some processing in the middle should be long, otherwise there will be a lack of precision.

Code:
[Cpp]
# Include <algorithm>
# Include <iostream>
# Include <cstring>
# Include <string>
# Include <vector>
# Include <cstdio>
# Include <stack>
# Include <queue>
# Include <cmath>
# Include <set>
Using namespace std;
# Define maxn100000
 
Int t, n;
Long num [MAXN]; // total number of values that are saved
 
// Initialize the table
Void init (){
Long I, j, k, tmp; // long note
Memset (num, 0, sizeof (num); // Initialization
For (I = 1; I <= 5; I ++) {// The maximum number of enumerative digits.
For (j = pow (10, I-1); j <pow (10, I); j ++ ){
Num [j] = num [J-1]; tmp = (j-pow (10, I-1) + 1) * I;
For (k = 1; k <I; k ++)
Tmp + = (pow (10, k)-pow (10, k-1) * k;
Num [j] + = tmp;
}
}
}
 
Void solve (){
Int I, j, k, pos, ans;
// Locate the pos position
For (I = 1; I <MAXN; I ++ ){
If (n> num [I-1] & n <= num [I]) {
Pos = I; break;
}
}
// Search
Int cnt = n-num [pos-1]; int sum = 0;
Int len, tmp, tmp_j;
For (j = 1; j <= pos; j ++ ){
Tmp_j = j;
For (I = tmp_j, len = 0; I! = 0; I/= 10) len ++;
For (k = len-1; k> = 0; k --){
Ans = tmp_j/pow (10, k); sum ++;
If (sum = cnt ){
Printf ("% d \ n", ans );
Return;
}
Tmp = pow (10, k); tmp_j % = tmp;
}
}
}
 
Int main (){
// Freopen ("input.txt", "r", stdin );
Init (); scanf ("% d", & t );
While (t --){
Scanf ("% d", & n); solve ();
}
Return 0;
}