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 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
大意：从1开始排序，各个串为，1,12,123,1234,.....，12345678910，现在输入一个下标，让你输出这个下标的数是多少.
我们先开两个数组，一个数组存以该数为最大值的串其中有a[i]个数，另一个存以该数为最大值的串的最开始数的下标b[i]，那么输入该m，i从1开始，t+=a[i]，如果超过了，记录这个i值，即最大值，再把超过的减去原来的m就是该数的现在的第几位,那么只要输出该数除以有的位数再取余就是所求。

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
long long  a[50000],b[50000];
long long  i,t,m,tt;
long long pos ,ans;
int main()
{
    a[1] = 1;
    b[1] = 1;
    for( i = 2; i < 38000; i++){
        a[i] = a[i-1] + (int)log10((double)i) + 1;//数目
        b[i] = b[i-1]+a[i];}
    cin>>tt;
    while(tt--){
        cin >> m;
         i = 1;
        while(m > b[i]){
                i++;
        }
         pos = m - b[i-1];//据起始点的位数
         t = 0;
        for( i = 1;t < pos;i++)
        {
            t+=(int)log10((double)(i))+1;
        }//t为总数
        pos = t - pos;//pos就为该数的位数
        i--;//i为这个数
        ans = i/(int)pow(10.0,(double)pos)%10;
        printf("%lld
",ans);
    }
    return 0;
}