POJ1019 -- Number Sequence (large Number processing)

Number Sequence

Description
A single positive integer I is given. write a program to find the digit located in the position I in the sequence of number groups S1S2... sk. each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another.
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910
Input
The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. the line for a test case contains the single integer I (1 ≤ I ≤2147483647)
Output
There shoshould be one output line per test case containing the digit located in the position I.
Sample Input
2
8
3
Sample Output
2
2

Question:

Given a string, the structure is as follows:

1121231234... 123456789101112... 12345678910111213... N

Ask the number of I-bits in the string.

Solution:

Divides a string into N segments.

1 12 123 1234... 1234567891011... 123456789101112... N

Then for the length of the K segment len [k] = len [k-1] + K the length of the number. Len [k] = len [k-1] + log10 (k) + 1.

Then define sum [k] = sum [k-1] + len [k]. You can locate the field where I is located by comparing the size of sum [] and I.

Suppose I is in the k field (1234567... t... k), then I-sum [k-1] represents the position of I in the k field.

The formula log10 (j) + 1 can be used to determine the t where I is located and the position pos in t.

Ans = t/pow (10, pos) % 10.

Code:

1 /************************************** * *********************************** 2> File Name: poj1019.cpp 3> Author: Enumz 4> Mail: [email protected] 5> Created Time: 6 ******************************* **************************************** */7 8 # include <iostream> 9 # include <cstdio> 10 # include <cstdlib> 11 # include <string> 12 # include <cstring> 13 # include <list> 14 # include <queue> 15 # include <stack> 16 # include <map> 17 # include <set> 18 # include <algorithm> 19 # include <cmath> 20 # include <bitset> 21 # include <climits> 22 # define MAXN 4000023 using namespace std; 24 long len [MAXN], sum [MAXN]; 25 void init () 26 {27 for (int I = 1; I <MAXN; I ++) 28 {29 len [I] = len [i-1] + (int) log10 (double) I) + 1; 30 sum [I] = sum [i-1] + len [I]; 31} 32} 33 int solve (int N) 34 {35 int I = 1; 36 while (sum [I] <N) 37 I ++; 38 N-= sum [i-1]; 39 int len_k = 0, t; 40 for (t = 1; len_k <N; t ++) 41 len_k + = (int) log10 (double) t) + 1; 42 int pos = len_k-N; 43 return (T-1)/(int) pow (double) 10, pos) % 10; 44} 45 int main () 46 {47 int T; 48 cin> T; 49 int N; 50 init (); 51 while (T --) 52 {53 cin> N; 54 cout <solve (N) <endl; 55} 56 return 0; 57}

 

POJ1019 -- Number Sequence (large Number processing)