poj 1019 Number Sequence-优快云博客

本文介绍了一种算法，用于在由多个数字组组成的序列中查找特定位置的元素。每个组包含从1到某个最大值的一系列数字，数字按顺序连续排列。通过算法计算每个组的长度并定位目标位置，最终找到目标位置对应的数字。

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25686		Accepted: 6963

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

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    unsigned long a[50000];
int i,n,j,m;
    vector<int>v;
    a[0]=0;
    j=1;
for(i=1;i<50000;i++)
    {
        a[i]=a[i-1]+j;
if(i==9||i==99||i==999||i==9999)
            j++;
    }
    cin>>n;
while(n--)
    {
        scanf("%d",&m);
        i=0;
        v.clear();
while(m>a[i])
        {
            m-=a[i];
            i++;
        }
        i=0;
while(m>a[i])
            i++;
        m-=a[i-1];
while(i)
        {
            v.push_back(i%10);
            i/=10;
        }
        printf("%d\n",*(v.end()-m));
    }
}