[poj 1019] Number Sequence 数学 想法题

Number Sequence
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 37345 Accepted: 10777

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
Tehran 2002, First Iran Nationwide Internet Programming Contest

题目链接：http://poj.org/problem?id=1019
题意：有一串数字串，其规律为

1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011 123456789101112······k

输入位置n，计算这一串数字第n位是什么数字，注意是数字，不是数！例如12345678910的第10位是1，而不是10，第11位是0，也不是10。总之多位的数在序列中要被拆分为几位数字，一个数字对应一位。
思路：
(int)log10((double)i)+1

(i-1)/(int)pow((double)10,len-pos)%10 计算位置

代码：
s[i]:i组长度的前缀和；
a[i]：i组的长度
对任意一个数所占的空间很好求，即 （int）log10（k）+1；

#include<iostream>#include<stdio.h>#include<math.h>  #include<algorithm>using namespace std;long long a[31270],s[31270];void pre(){    a[1]=1;    s[1]=1;    for(int i=2;i<31269;i++)    {        a[i]=a[i-1]+(int)log10((double)i)+1;        s[i]=s[i-1]+a[i];    }}int cal(int n){    int i=0;    while(s[i]<n) i++;  //找在那组    int pos=n-s[i-1];  //找在那位置    int len=0;    int j;    for(j=1;len<pos;j++)    {         len+=(int)log10((double)j)+1;      }     //找在那个数    return ((j-1)/(int)pow(10.0,len-pos))%10;      //在i数的len-pos位置的数字；}int main(){    pre();    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        printf("%d\n",cal(n));    }}