POJ 1019：Number Sequence 二分查找

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36013		Accepted: 10409

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

Sample Output

2

2

题意是给定了一串有规律的数，问位置i的数是0~9中的哪一个。

自己一直再跟着POJ分类的题目在做，然后很自然地就会总看小优的博客。。。但是觉得这道题没有小优说的那么夸张，多么难。

首先计算每一个数的长度，比如11，n[11]=2。

之后计算从1到n 这一段数的长度，比如1234567891011，所以c[11]=13。

最后计算总体的长度，比如1121231234123451234561234567123456781234567891234567891012345678910111，即s[11]=70

初始化这些结束之后，剩下的就是不断切分，先二分找s数组，之后定位到哪一段，即是c数组的事情，在之后定位到是哪一个数，是n数组的事情，再然后就是这个数的第几位，手动算吧。这样逐级下来，就能得到第几个数是什么。

主要就是做的时候思路清晰，别乱就好。

代码：

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

#include <cstring>

#pragma warning(disable:4996)

using namespace std;

int c[52850];

long long s[52850];

int n[52850];

int num;

int length(int x)

{

	int sum=0;

	while (x != 0)

	{

		x /= 10;

		sum++;

	}

	return sum;

}

void init()

{

	int i;

	c[0] = 0;

	s[0] = 0;

	for (i = 1; i <= 52849; i++)

	{

		n[i] = length(i);

	}

	for (i = 1; i <= 52849; i++)

	{

		c[i] = c[i - 1] + n[i];

	}

	for (i = 1; i <= 52849; i++)

	{

		s[i] = s[i - 1] + c[i];

	}

}

int main()

{

	init();

	int Test,left,pos;

	int i;

	scanf("%d", &Test);

	while(Test--)

	{

		scanf("%d", &num);

		left = lower_bound(s + 1, s + 52845, num) - (s + 1);

		pos = num - s[left];

		left= lower_bound(c + 1, c + 52845, pos) - (c + 1);

		pos = pos - c[left];

		left++;

		pos = n[left] - pos +1;

		i = 1;

		while (i < pos)

		{

			left /= 10;

			i++;

		}

		cout << left % 10<<endl;

	}

	return 0;

}

巴特西

POJ 1019：Number Sequence 二分查找

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