poj 1019 Number Sequence

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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

283

Sample Output

22

#include <iostream>#define MAX 31269#include<cmath>using namespace std;int a[MAX];long long  b[MAX];int main(){    a[1]=1;    b[1]=1;    for(int i=2; i<MAX; i++ )    {        a[i]=a[i-1]+ (int )log10((double)i)+1;        b[i]=b[i-1]+a[i];        //log10(i)+1 表示第i组数字列的长度 比 第i-1组 长的位数        //前i组的长度s[i] 等于 前i-1组的长度s[i-1] + 第i组的长度a[i]    }    //log()是重载函数，必须对int的i强制类型转换，以确定参数类型    int n;    int t;    cin>>t;    while(t--)    {        cin>>n;        int i=1;        while(b[i]<n)        {            i++;        }        int pos=n-b[i-1];        int len=0;        for(i=1; len<pos; i++)        {            len+=(int )log10((double)i)+1;        }        cout<<(i-1)/(int )pow((double)10,len-pos)%10<<endl;        //之所以i-1，是因为前面寻找第i组长度时，i++多执行了一次        //i=i-1 此时i刚好等于第n位个置上的数 （数是整体，例如123一百二十三，i刚好等于123，但n指向的可能是1，2或3）        //pos为n指向的数字在第i组中的下标值        //len为第i组的长度        //那么len-pos就是第i组中pos位置后多余的数字位数        //则若要取出pos位上的数字，就要利用(i-1)/pow(10,len-pos)先删除pos后多余的数字        //再对剩下的数字取模，就可以得到pos        //例如要取出1234的2，那么多余的位数有2位：34。那么用1234 / 10^2，得到12，再对12取模10，就得到2    }    return 0;}