Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second

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…S_k. Each group S_k 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 <=25), 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                           Output for Sample Input

Problem source: Iranian Contest

Special Thanks: Shahriar Manzoor, EPS.

思路：这道题暴力肯定超时，我想了很久，没想到什么方法，后来参照了他人的思路和代码。下面是题解

观察数字可以找到规律：

数字范围          数字位数范围    每个数字宽度   总共数字所占位数

1~9 ：              1~9.                     1                      45

10~99：           11~189                 2                      9000

100~999：       192~2889              3                     1386450

10000~9999:    2893~38889          4                     1881019000  

>=100000         38894 ~ *              5                      **

找出输入数字所在的数字位数范围，再进一步确定他具体所在的数字位数，再进行枚举。

比如说：输入 9045，由上表先可以确定它是在11~189数字位数范围内，再进一步可以确定他是在189（即：从1到99）这个位数范围里的，最后进行枚举，发现他恰好189这个位数范围内的数字99的个位上，因此输出 9 。

代码：

#include<iostream>
#include<cmath>
#include<cstdio>

using namespace std;


void solve(int num,int finish,int start,int len)
{
    int j;
    long long i,sum=0;
    char ch[6];
    for(i=0;i<finish;start+=len)
    {
        i=i+start;
        if(num<=i)
        {
            i=i-start;
            break;
        }
    }
    int len1,cnum=num-i;
    for(j=1;;j++)
    {
        len1=(int)log10(j)+1;
        sum=sum+len1;
        if(sum>=cnum) break;
    }
    sum-=len1;
    sprintf(ch,"%d",j);
    cout<<ch[cnum-sum-1]<<endl;
    return;
}


int main()
{
    int n,m;
    cin>>n;
    while(n--)
    {
        cin>>m;
        if(m<=45)
            solve(m,45,1,1);
        else if(m<=9000+45)
        {
            m=m-45;
            solve(m,9000,11,2);
        }
        else if(m<=9000+45+1386450)
        {
            m=m-9045;
            solve(m,1386450,192,3);
        }
        else if(m<=1386450+9045+188019000)
        {
            m=m-9045-1386450;
            solve(m,188019000,2893,4);
        }
        else
        {
            m=m-188019000-9045-1386450;
            solve(m,2147483647,38894,5);
        }
    }
    return 0;
}

[模拟]uva10706 - Number Sequence,布布扣,bubuko.com

[模拟]uva10706 - Number Sequence

原文：http://blog.csdn.net/zju_ziqin/article/details/21040625