UVa 10706/poj 1019 number Sequence: Play table and O (1) algorithm

10706-number Sequence

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=19&page=show_problem &problem=1647

http://poj.org/problem?id=1019

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 integers numbers ranging from 1 to K, written one after another.
For example, the digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The ' 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 should be one output line per test case containing the digit located in the position I.

Sample Input

2
8
3

Sample Output

2
2

Source

Tehran 2002, Iran Nationwide Internet programming Contest

Complete code:

Complexity: O (n), but the constant term is small

/*uva:0.016s*/
/*poj:0ms,648kb*/
  
#include <cstdio>
#include <cmath>
  
int a[31270], s[ 31270];
  
inline int pow_10 (int x, int d)
{while
    (d--) x/=;
    return x%;
}
  
int main (void)
{
    int t, N, T, I;
    for (i = 1;; i++)
    {
        A[i] = a[i-1] + (int) (LOG10 (double) i) + 1;
        S[i] = S[i-1] + a[i];
        if (S[i] < 0) break;
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d", &n);
        i = 0;
        while (S[i] >= 0 && s[i] < n) i++;///n the group
        t = n-s[i-1];
        i = 0;
        while (A[i] < T) i++;///n the single-digit number of digits
        printf ("%d\n", pow_10 (i, a[i)-T);
    return 0;
}

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/