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
解决方案：
对于这题，我们要用到math.h中的log10函数。在这里，我们有个规律，对于每一个正整数value，其位数等于log10(value) + 1.
我们用一个数组a来保存每个子串的位数（如子串123456789101112的位数为15）.
我们用一个数组b来保存每个子串以及此子串之前所有的位数。
由于i有其取值范围，故我们可以选择数组维数大小为32000
下面就开始寻找特定位置上的那个数了。
1.首先根据所输入的位置来确定在哪个子串中。
2.确定是字串的哪个数。
3.再确定是该数的第几位。
4.然后输出之。
具体代码如下(gcc编译时加上-lm参数)：
POJ 1019
1 #include <stdio.h> 2 #include <time.h> 3 #include <math.h> 4 #include <stdlib.h> 5 //log10(n)就可以求出n的位数 6 #define LEN 32000 7 unsigned int a[LEN]; 8 unsigned int b[LEN]; 9 int main() 10 { 11 a[0] = 0; 12 a[1] = 1; 13 b[0] = 0; 14 b[1] = 1; 15 int i; 16 for(i = 2; i < LEN; i++) 17 { 18 a[i] = a[i-1] + log10((double)i) + 1; 19 b[i] = b[i-1] + a[i]; 20 } 21 22 unsigned int cases; 23 unsigned int n; 24 scanf("%u", &cases); 25 while(cases--) 26 { 27 scanf("%u", &n); 28 unsigned int j; 29 for(j = 0; j < LEN; j++) 30 if(n <= b[j]) 31 break; 32 //在字串中的位置 33 unsigned int remain = n - b[j-1]; 34 35 unsigned int value; 36 unsigned int sum = 0; 37 for(value = 1; value < LEN; value++) 38 { 39 sum += 1+log10((double)value); 40 if(sum >= remain) 41 break; 42 } 43 //value的位数 44 unsigned int value_bits = 1+log10((double)value); 45 unsigned int position = remain - (sum - value_bits); 46 //求value的第position位 47 char* str = (char*)malloc(10*sizeof(char)); 48 int temp = value; 49 int stri = 0; 50 while(temp) 51 { 52 int temp1 = temp%10; 53 str[stri++] = temp1 + '0'; 54 temp = temp/10; 55 56 } 57 str[stri] = '\0'; 58 if(position == value_bits) 59 printf("%c\n", str[0]); 60 else 61 printf("%c\n", str[position - 1]); 62 } 63 return 0; 64 }