Uva10706-number sequence (query rule)

Question: uva10706-number sequence (query rule)

There is such a series of 11212312341234512345612345671234567812345678912345678910123456789101112345678910..., ask the value of the I position.

1 2 3 4 5 6 7...

Solution: the rule needs to be discovered. At the beginning, I still read the wrong question. The rule can be understood only after reading the answers of others.

Here we define s [I] to represent the position of the I-th sequence. For example, the numbers below the string sequence represent the sequence to which the string belongs. (Position 1, 3, and 6 ...)

Num [I] indicates the length of the I-th sequence. Note that the number is 10 characters long.

Recursive Formula: s [I] = s [I-1] + s [I-1]-s [I-2] + wi (number of digits of I ).

To find the position of the I sequence, we need to start from the position of the I-1 sequence + the length of the I sequence. S [I-1]-s [I-2]: The position of the I-1 sequence, minus the position of the I-2 sequence, is the length of the I-1 sequence, that is, num [I-1 ]. Then, as long as the length of I-1 is added to the position of the I-1 sequence, the length difference between the I sequence and the I-1 sequence is the number of I digits.

Here we need to pre-process the arrays S and num, and then find the corresponding values by traversing these two arrays. The size of the array can be increased to 1 E5, which is more than 2147483647.

To find the value at the position I, you must first find the sequence in which it is located. Traverse the S array and find s [k-1] <= I. Subtract I from S [k-1] (position of the k-1 sequence) it is the length of the I position in the K sequence.

Then the length of each sequence also follows the regular rule: sequence 3: 123 length 3

Sequence 4: 1234 length 4

The length difference between sequence I and sequence I-1 is actually the number of digits of sequence I, and it is incremental.

Find num [k-1]> = I <= num [k] In the num array, which means that the value of I is one of the K digits.

Code:

# Include <stdio. h> # include <math. h> typedef long ll; const ll n = 2147483647; const int W [] = {0, 10,100,100 0, 10000,100 000 }; // (0, 1, 2 ...) number of BITs dividing value const int M = 100000; ll s [m]; ll num [m]; // pre-processing void Init () {s [0] = num [0] = 0; s [1] = num [1] = 1; int cur = 1; for (INT I = 2; s [I-1] <n; I ++) {if (I> = W [cur]) cur ++; s [I] = 2 * s [I-1]-s [I-2] + cur; num [I] = s [I]-s [I-1] ;}} int main () {Init (); int t; ll N; scanf ("% d", & T); char STR [10]; while (t --) {scanf ("% LLD", & N); int I; for (I = 1; I <m; I ++) if (s [I]> = N) break; n-= s [I-1]; for (I = 1; I <m; I ++) if (Num [I]> = N) break; n-= num [I-1]; sprintf (STR, "% d", I); printf ("% C \ n", STR [n-1]);} return 0 ;}