POJ 1019, Number Sequence

Time Limit: 1000MS  Memory Limit: 10000K
Total Submissions: 17288  Accepted: 4563

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

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//  POJ1019.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < sstream >
using   namespace  std;
inline  int  getlen( int  i)
{
     if  (i  <   10 )  return   1 ;
     else   if  (i  <   100 )  return   2 ;
     else   if  (i  <   1000 )  return   3 ;    
     else   if  (i  <   10000 )  return   4 ;    
     else   if  (i  <   100000 )  return   5 ;
     return   0 ;
}
inline  char  getchar( int  num,  int  p)
{
    stringstream ss;
    ss  <<  num;
     string  s;
    ss  >>  s;
     return  s[p - 1 ];
}
int  main( int  argc,  char *  argv[])
{
     // init table
     const   int  SIZE  =   40000 ;
    __int64 sum[SIZE];
    sum[ 0 ] = 0 ;

     for ( int  i  =   1 ;i  <  SIZE; ++ i)
        sum[i] = sum[i - 1 ] + getlen(i);

     for ( int  i  =   1 ; i  <  SIZE;  ++ i)
        sum[i]  += sum[i - 1 ];

     int  cases;
    scanf( " %d " ,  & cases);
    __int64 num;
     for  ( int  c  =   0 ; c  <  cases;  ++ c)
    {
        scanf( " %I64d " ,  & num);

         //  Kth group
         int  k  =   1 ;
         while (sum[k]  <  num)  ++ k;
        
         // posth character of number i
         int  pos  =  num  -  sum[k - 1 ];
         int  i = 1 ;
         while (pos - getlen(i) >   0 ) pos  -=  getlen(i),  ++ i;

        cout  <<  getchar(i,pos)  <<  endl;
    }

     return   0 ;
}