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
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
思路：我们这样来看这个数列：
1
12
123
1234
12345
..........................
用一个数组len来保存每一个1 to i 的总位数，这样就有了一个递推公式：len[i]=len[i]=len[i-1]+(int)log10(i)+1。(int)log10(i)+1可求i的位数。
接着，假设求第n位的数字，我们就令i=1; 如果len[i]<n就n-=n-=len[i++]，如此循环。 最后问题就转化成了求数列 1 to i 的第n位。
再另i=1，如果(int)log10(i)+1<n就n-=(int)log10(i)+1; i++;继续循环。 问题最终就转化成了求i的第n位数。 剩下的不用说了吧。
#include <stdio.h>
#include <math.h>
int len[32000];
int main()
{
int t,n,i,sum=0;
for(i=1;sum>=0;i++)
{
len[i]=len[i-1]+(int)log10(i)+1;
sum+=len[i];
}
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
i=1;
while(len[i]<n) n-=len[i++];
i=1;
while((int)log10(i)+1<n)
{
n-=(int)log10(i)+1;
i++;
}
i/=pow(10,(int)log10(i)+1-n);
i%=10;
printf("%d\n",i);
}
}