POJ 1019 Number Sequence (循环递增序列的的第K个值)

                                                                                                        Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35823		Accepted: 10340

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

题目大意：找出第i位的数值

思路：可以把序列分成若干组如1  12  123   1234  ....利用s[]a[]两个数组分别记录全部每序列的长度，和当前每组序列的长度，然后大体的找到i在哪一组，然后再在每一个

小的组里找到相应位置的这个值。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<cmath>
#include<queue>
#define LL long long
using namespace std;
LL s[100000],a[100000];
LL so(LL x,LL d)
{
    while(d--)
    {
        x/=10;
    }
    return x%10;
}
int main()
{
    LL n,m,i,j,cla;
    a[1]=s[1]=1;
    for(i=2;i<=34000;i++)//先打表
    {
        a[i]=a[i-1]+(LL)log10((double)i)+1;
        s[i]=s[i-1]+a[i];
    }
    scanf("%lld",&cla);
    while(cla--)
    {
        scanf("%lld",&n);
        i=0;
        while(s[i]<n)
            i++;
        m=n-s[i-1];//找到在n在全部序列里的下标
        i=0;
        while(m>a[i])
        i++;//i为所在n组的长度
        printf("%lld\n",so(i,a[i]-m) );//<span id="transmark"></span>例如要取出1234的2，那么多余的位数有2位：34。那么用1234 / 100，得到12，再对12取模10，就得到2</span>
    }
    return 0;
}