http://poj.org/problem?id=1019

题目：

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

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest
 
数学类问题，题意是给一串 1 12 123 1234 12345 123456 。。。。这样的数字问第

i个数字是多少。这题的trick在于第i个数与第i-1个数的位数差不定，可以利用公式:(int)log10(double(i))+1，当增加的数字是10的幂次关系时个数会变化。

打表：一开始没有打表，超时了~~~~呵呵····终于知道什么叫打表了！！

#include<math.h>
#include<iostream>
using namespace std;
unsigned int a[31270],s[31270];
void reset()//打表
{
    int i;
    a[1]=1;
    s[1]=1;
    for(i=2;i<31270;i++)
    {
        a[i]=a[i-1]+(int)log10((double)i)+1;
        s[i]=s[i-1]+a[i];
    }    
}    

int main()
{
    int T;
    int n;
    int i;
    scanf("%d",&T);
    reset();
    while(T--)
    {
        scanf("%d",&n);
        i=1;
        
        while(s[i]<n) i++;
      
        int pos=n-s[i-1]; 
        int tmp=0;
        for(i=1;tmp<pos;i++)
        {
            tmp+=(int)log10((double)i)+1;
        }   
        int k=tmp-pos;
        printf("%d\n",(i-1)/(int)pow(10.0,k)%10) ;/*从右向左求，比如123456，k=2,则结果为4*/

    }  
    return 0; 
     
}  

这道题是比较简单的了~~~~~~