Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31455		Accepted: 8924

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: 
1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011 12345678910

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
题意：已知题目中按一定规律排列的序列，输入i求第i个位置上对应的数字；

若干个小时的成果，记录一下思考过程。
思路：  将前2147483647个数分为5个区，
       第i个区     区间范围（num） 前num个数序列个数总和
        1         1——9          45
        2         10——99        9000
        3         100——999      1386450
        4         1000——9999    188019000
        5         10000——99999  2147483647
       即第i个区有i位数；
        
       再找第i个区的第j个数
       第i区的第一个数       该数对应序列的长度            
         1                   1
         10                  11
         100                 192
         1000                2893
         10000               38894
      
       再找第j个数的第k段    
       第k段            第k段包含的序列总长度
         1                9
         2                180
         3                2700
         4                36000
         5                450000
       找到第k段后就可以确定第i个数对应的十进制数了，例如求第80个位置时，它在第2个区的第3个数（十进制是12）的第2段，就可以确定第80个数对应的十进制数是10，然后输出第80位数字。

#include<stdio.h>
#include<string.h>
int x;
int seq_sum[] = {0,45,9000,1386450,188019000,2147483647};
int seq_len[] = {0,1,11,192,2893,38894};
int seq_num[] = {0,1,10,100,1000,10000};
int seq_part[] = {0,9,180,2700,36000,450000};

void solve()
{
    int i,j,k,t,asi,s;
    for(i = 1; ; i++)//第i区；
    {
        if(x > seq_sum[i])
            x = x-seq_sum[i];
        else break;
    }
   
    for(j = 1; ; j++)//第j个数;
    {
        int tmp = seq_len[i]+(j-1)*i;
        if(x > tmp)
            x = x-tmp;
        else break;
    }
  

    for(k = 1; ; k++)//对应第k段;
    {
        if(x > seq_part[k])
            x = x-seq_part[k];
        else break;
    }

    t = x/k;
    asi = x%k;
   
    if(asi != 0) t++;
    else asi = k;
  
    s = seq_num[k]+t-1;

    for(int l = 1; l <= k-asi; l++)
        s/=10;
    printf("%d
",s%10);
}

int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d",&x);
        solve();
    }
    return 0;
}