E - Number Sequence（第二季水）

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

这道题不打表会超时 ，以下 超时代码

#include<iostream>
#include<cmath>
using namespace std;
void f(int n)
{
    int p=,j=,t;
    bool flag;
    while(){
        flag=false;
        for(j=;j<=p;j++){
            int i;
            for(i=;;i++){
                t=pow((double),i);
                if(j/t==){
                    n-=i;
                    break;
                }
            }
            if(n<=){
                int k=i+n;
                flag=true;
                n=j%(int)(pow((double),i-k+))/pow((double),i-k);
                break;
            }
        }
        if(flag==true)break;
        p++;
    }
    cout<<n<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        f(n);
    }
    //system("pause");
    return ;
}

打表后！！！这个代码不好理解

#include<iostream>
#include<cmath>
using namespace std;
unsigned int a[],s[];
void f()
{
    int i;
    a[]=;
    s[]=;
    for(i=;i<;i++)
    {
        a[i]=a[i-]+(int)log10((double)i)+;   //记录1至s[i]个数字的位数和
        s[i]=s[i-]+a[i];                      //一位 记录 1至s[i]个数字
    }
}
int main()
{
    int t;
    int n;
    int i;
    cin>>t;
    f();
    while(t--)
    {
        cin>>n;
        i=;
        while(s[i]<n) i++;
        int pos=n-s[i-];
        int tmp=;
        for(i=;tmp<pos;i++)           //第n个数字在s[i-1]这个数据组中
        {
            tmp+=(int)log10((double)i)+;
        }
        int k=tmp-pos;       //数字i从低位数的第k+1位
        cout<<(i-)/(int)pow(10.0,k)%<<endl;
    }
    return ;
}