E - 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


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=1,j=1,t;
    bool flag;
    while(1){
        flag=false;
        for(j=0;j<=p;j++){
            int i;
            for(i=0;;i++){
                t=pow((double)10,i);
                if(j/t==0){
                    n-=i;
                    break;
                }
            }
            if(n<=0){
                int k=i+n;
                flag=true;
                n=j%(int)(pow((double)10,i-k+1))/pow((double)10,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 0;
}

 

打表后！！！这个代码不好理解

 

#include<iostream>
#include<cmath>
using namespace std;
unsigned int a[31270],s[31270];
void f()
{
    int i;
    a[1]=1;
    s[1]=1;
    for(i=2;i<31270;i++)
    {
        a[i]=a[i-1]+(int)log10((double)i)+1;   //记录1至s[i]个数字的位数和
        s[i]=s[i-1]+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=1;
        while(s[i]<n) i++;
        int pos=n-s[i-1]; 
        int tmp=0;
        for(i=1;tmp<pos;i++)           //第n个数字在s[i-1]这个数据组中
        {
            tmp+=(int)log10((double)i)+1;
        }   
        int k=tmp-pos;       //数字i从低位数的第k+1位
        cout<<(i-1)/(int)pow(10.0,k)%10<<endl;
    }  
    return 0; 
}