POJ 1019 数学

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36482 Accepted: 10525

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<cstdio>#include<iostream>#include<cmath>using namespace std;unsigned int a[31270]={0},s[31270]={0};int main(){ a[1]=s[1]=1; for(int i=2;i<31270;i++)    {     a[i]=a[i-1]+(int)log10((double)i)+1;        s[i]=s[i-1]+a[i]; } int T=0; scanf("%d",&T); while(T--) {  int n=0;  scanf("%d",&n);  int i=1;  while(s[i]<n) i++;  int pos=n-s[i-1];  int tem=0;  i=1;  for(i=1;tem<pos;i++)  {   tem+=(int)log10((double)i)+1;  }  pos=tem-pos;  printf("%d\n",(i-1)/(int)pow(10.0,pos)%10); } return 0;}