POJ 1019 Number Sequence

找规律，先找属于第几个循环，再找属于第几个数的第几位。。。。。。

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31552		Accepted: 8963

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 

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

using namespace std;

inline int getlen(int x)
{
    return log10(1.0*x)+1;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int pos,th,kth=0,x=1,nth=0,i;
        scanf("%d",&pos); th=pos;
        while(pos>0)
        {
            kth+=getlen(x);
            pos-=kth;
            x++;
        }
        x=x-1; pos=pos+kth;
        for(i=1;i<=x;i++)
        {
            nth+=getlen(i);
            if(nth>=pos)
                break;
        }
        nth-=getlen(i);
        int deta=pos-nth;
        int bit[10],ii=0;
        while(i)
        {
            bit[ii++]=i%10;;
            i/=10;
        }
        printf("%d ",bit[ii-deta]);
    }
    return 0;
}

/*   有爱的测试数据。。。
            1
            1
            2
            1
            2//5
            3
            1
            2
            3
            4//10
            1
            2
            3
            4
            5//15
            1
            2
            3
            4
            5//20
            6
            1
            2
            3
            4//25
            5
            6
            7
            1
            2//30
            3
            4
            5
            6
            7//35
            8
            1
            2
            3
            4//40
            5
            6
            7
            8
            9//45
            1
            2
            3
            4
            5//50
            6
            7
            8
            9
            1//55
            0
            1
            2
            3
            4//60
            5
            6
            7
            8
            9//65
            1
            0
            1
            1
            1//70
            2
            3
            4
            5
            6//75
            7
            8
            9
            1
            0//80
*/

* This source code was highlighted by YcdoiT. ( style: Codeblocks )