poj 1019Number Sequence(数学巧妙~~~)

最新推荐文章于 2022-12-03 10:34:43 发布

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本文介绍了一个算法问题，即在一个特殊的正整数序列中找到第i位上的数字。该序列由多个组Sk构成，每组包含从1到k的连续正整数。通过预先计算和存储关键数据，可以高效地解决这一问题。

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37890		Accepted: 10952

Description

A single positive(积极的) integer(整数) i is given. Write a program to find the digit(数字) located(处于) in the position i in the sequence(序列) 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(投入) file contains a single integer(整数) 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(输出) line per test case containing the digit(数字) located(处于) in the position i.

Sample Input

2
8
3

Sample Output

2
2

#include<stdio.h>
#include<math.h>
unsigned int a[32000];
unsigned int s[32000];
void play_table()
{
    int i,j;
    a[1]=1;
    s[1]=1;
    for(i=2;i<32000;i++)
    {
        a[i]=a[i-1]+(int)log10((double)i)+1;
        s[i]=s[i-1]+a[i];

    }
}
int slove(unsigned n)
{
    int i=1;
    while(n>s[i]) i++;

    int pos = n-s[i-1];
    int len = 0;
    for(i=1;len<pos;i++)
        len+=(int)log10((double)i)+1;
    return (i-1)/(int)pow(10.0,len-pos)%10;
}
int main()
{
    unsigned n;
    int t,i,j;
    play_table();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%u",&n);
        printf("%d\n",slove(n));
    }
}