Number Sequence


Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 38785

 

Accepted: 11258


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

题意：有这么一个序列1 12 123 1234 12345 ... 1234567891011... 输入一个i；输出第i个数是什么。

题解：step1：分组，用到(int)log10((double)i)+1；

step2：找到i所在的组；

step3：求出pos所对应的位置。

#include <iostream>
#include <math.h>
#include <stdio.h>

using namespace std;

int a[31270];
long long b[31270];
void init()//打表分组。
{
    int i;
    a[0] = 0;
    b[0] = 0;
    for(i = 1; i < 31270; i++)
    {
        a[i] = a[i-1] + (int)log10((double)i) + 1;
        b[i] = b[i-1] + a[i];
    }
//    for(i = 0; i < 3127; i++)
//    {
//        cout<<i<<" "<<a[i]<<" "<<b[i]<<endl;
//    }
}

int main()
{
    init();
    int K, i;
    int n;
    cin>>K;
    while(K--)
    {
        cin>>n;
        i = 0;
        while(n > b[i])
        {
            i++;
        }
        int pos = n - b[i-1];//找到所在组数，并找到所在组的第几个；
        int s = 0;
        for(i = 1; s < pos; i++)
        {
            s += (int)log10((double)i) + 1;
        }//得到pos是第i-1个数的一部分。
        int k=s-pos;//找到是i-1的第几个；
        printf("%d\n",(i-1)/(int)pow(10.0,k)%10);//从左到右找到
    }
    return 0;
}