Number Sequence
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24387 | Accepted: 6534 |
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
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
题目大意就是给你这一串数字11212312341234512345612345671234567812345678912345678910123456789101112345678910……（未列完）
要我们求出第n个数是多少(从左到右看)，例如第2个是1，第三个是2，第八个是2；
如果仔细观察这一串数字，可以发现他可以还分为很多小串，假设第i小串是123……i，假设第i小串所占的空间是a[i]，则通过对比a[i]与a[i+1]发现，
第i+1串只比第i串多一个数，即i+1，故他们所占的空间差就是第i+1所占的空间。
对任意一个数所占的空间很好求，即 （int）log10（k）+1；
然后就可以求出每一个串的起始位置，通过与n比较就可以确定n出现在那一个串里，最后在求出n在这个串里的相对位置，就可以求出该题的解
参考代码
1 #include<iostream>
2 #include<cmath>
3 #include<cstdio>
4 using namespace std;
5 int a1[32000] ; //用来存储每一个串所占的空间
6 __int64 a[32000] ; //用来存储每一个串的起始位置
7 int num[150000]; //打印出最大的一个串
8 int main()
9 {
10 int i;
11 a1[0]=0 ;
12 a1[1]=1;
13 for ( i = 2 ; i < 32000 ; i ++ )
14 a1[i] = a1[i-1] + (int)log10(1.0*i)+1; //推导详见上文
15 a[0]=1;
16 for ( i = 1 ; i < 32000 ; i ++ )
17 a[i]=a[i-1]+a1[i-1]; //上一个串的起点加所占空间就是下一个串的起点
18 int k = 1;
19 for ( i = 1 ; i < 31300 ; i ++ )
20 {//打印最大的一个串
21 char str[20];
22
23 str[0]='0';
24 int ti = i ;
25 int len = 0 ;
26 while (ti)
27 {
28 str[len]=ti%10+'0';
29 ti=ti/10;
30 len++;
31 }
32 while ( len -- )
33 {
34 num[k]=str[len]-'0';
35 k++;
36 }
37 }
38 int t;
39 cin >> t;
40 while ( t-- )
41 {
42 int mn;
43 cin >>mn;
44 for ( i =1; i < 32000 ; i ++ )
45 if ( a[i]>=mn )
46 break;
47 if ( a[i] ==mn )
48 cout<<1<<endl;
49 else
50 {//mn-a[i-1]+1就是其在第i-1个串里的相对位置
51 cout<<num[mn-a[i-1]+1]<<endl;
52 }
53 }
54 return 0;
55 }