poj Number Sequence

Number Sequence （排列组合）

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

Submit Status

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

1. #include<iostream>  
2. #include<cstdio>  
3. #include<cmath>  
4. #define max 31270  
5. using namespace std;  
6. unsigned int a[max];  
7. unsigned int s[max];  
8. void start()  
9. {  
10.     int i,j;  
11.     a[1]=1;  
12.     s[1]=1;  
13.     for(i=2; i<31270; i++) //打表记录总的位数  
14.     {  
15.         a[i]=a[i-1]+(int)log10((double)i)+1;  
16.         s[i]=s[i-1]+a[i];  
17.     }  
18. }  
19. int res(int n)//预处理，不断缩小范围  
20. {  
21.     int i=1,j=1;  
22.     int m=0;  
23.     int len=0;  
24.   
25.     for(i=1; i<31270; i++)  
26.     {  
27.         if(s[i]>=n)  
28.           break;  
29.     }  
30.     int pos=n-s[i-1];  
31.     for(j=1; len<pos; j++)  
32.     {  
33.         len+=(int)log10((double)j)+1;  
34.     }  
35.     return ((j-1)/(int)pow(10.0,len-pos))%10;  
36. }  
37. int main()  
38. {  
39.     int t;  
40.     int test;  
41.     cin>>t;  
42.     start();  
43.     while(t--)  
44.     {  
45.         cin>>test;  
46.         cout<<res(test)<<endl;  
47.     }  
48.     return 0;  
49. }