poj1019——Number Sequence（数学）

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

给出了一个实际上是1 12 123 1234….的序列 ，求第n个位置上的数是多少。
由于10，100这样的多位数占了不止一个位置，所以要计算长度，具体的可以看题解http://blog.csdn.net/lyy289065406/article/details/6648504 ，每一行代码都值得思考，需要都看懂，反正我是看了很久。
题解中的注释里应该有一处错误，我在自己的代码里写了出来。

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <map>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 20005#define Mod 10001using namespace std;unsigned int a[31270],s[31270];void reset(){    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,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++)  //这个i应该是在之前得出的第i组里，从1开始遍历得到pos所在位的数字，并非题解中的第i组        {            tmp+=(int)log10((double)i)+1;        }        printf("%d\n",(i-1)/(int)pow(10.0,tmp-pos)%10);    }    return 0;}