UVa 10706 / POJ 1019 Number Sequence (打表&O(1)算法)

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 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 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, First Iran Nationwide Internet Programming Contest

完整代碼：

複雜度：O(n)，但常數項很小

/*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 /= 10;
	return x % 10;
}

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所在的組
		t = n - s[i - 1];
		i = 0;
		while (a[i] < t) i++;///n所指向的數的個位數
		printf("%d\n", pow_10(i, a[i] - t));
	}
	return 0;
}
           

O(1)代碼見 此。