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
這道題目其實就是一道數學歸納題，用遞歸方法可以求得其結果，爲此需要如下參數：
Len[N]：表示第一次出現數字Ｎ(而不是字符串N)時，數字序列的長度。
執行結果：Calc[N]：表示數字序列123…N的數字個數。
對這道題目必定有：
假設要求的數字序列的第N個數字idata時，其算法如下：
Step 1: 求出小於等於2147483647的所有Len以及相應的Calc；
Step 2: 尋找Len[i]使得Len[i]<=N<Len[i+1];
若Len[i]=N，則idata=i%10; 算法結束。
否則(Len[i]<N<Len[i+1]), N-=Len[i]；
Step 3: 尋找Calc[i]使得Calc[i]<=N<Len[i+1];
若Calc[i]=N，則idata=i%10；算法結束。
否則(Calc[i]<N<Calc[i+1])，N-=Len[i], idata = i+1的第N位數；算法結束。
具體實現如下：
#include "iostream"
#include "math.h"
using namespace std;
const int N = 100000;
unsigned int Len[N];
unsigned int Calc[N];
void LenAndLen(void)
{
int i = 1;
Len[1] = 1;
Calc[1]= 1;
while(Len[i++] < 2147483647)
{
Calc[i] = Calc[i-1] + (int)floor(log10((double)i) + 1.0);
Len[i] = Len[i-1] + Calc[i];
}
}
int SearchData(unsigned int N)
{
int i = 1;
while(Len[i] < N)
{
i++ ;
}
if(N == Len[i])
{
return i%10;
}
else
{
N -= Len[i-1];
i = 1;
while(Calc[i] < N)
{
i++ ;
}
if(N == Calc[i])
{
return i%10;
}
else
{
N -= Calc[i-1];
return i/((int)pow(10., (double)((int)log10((double)(i*10)))-N))%10;
}
}
}
int main(void)
{
unsigned int t, n;
LenAndLen();
cin >> t;
while(t --)
{
cin >> n;
cout << SearchData(n) << endl;
}
return 0;
}