POJ 1019 Number Sequence【数论】

转自 http://www.cnblogs.com/Hilda/archive/2012/08/17/2644739.html

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题目大意：比较容易理解就是找出第i位的数字是多少；思路：

很容易看出题目给的规律，1 112 112123 1121231234 112123123412345....题目给出第i位的数字，i的范围恰好是int的范围，可以算出不会超过40000组；
用a[i]表示i组数的位数：a[i]=a[i-1]+(int)(log10(i*1.0))+1; num[i]表示前i组数的位数和num[i]=num[i-1]+a[[i];
举例：假若求54321的第2位的数字，即54321/(pow(10, 5-2))%10;即将多余的位数去掉：n/(pow(10, 多余位数），在对10取模得打结果；

关于精度问题解题思路没有精确说明；

代码如下：

#include<stdio.h>#include<string.h>#include<math.h> long long a[40005], num[40005]; int main(){    int T, i;    long long n;     scanf("%d", &T);    memset(a, 0, sizeof(a));    memset(num, 0, sizeof(num));     for(i=1; i<=40000; i++)    {        a[i]=(long long)(log10(i*1.0))+1+a[i-1];        num[i]=num[i-1]+a[i];    }    while(T--)    {        scanf("%lld", &n);        for(i=0; i<40000; i++)            if(n>num[i]&&n<=num[i+1])                break;         int pos=n-num[i];        int length=0;        for(i=1; length<pos; i++)            length+=(int)(log10(i*1.0))+1;         int mod=(i-1)/((int)(pow(10.0, length-pos)))%10;        printf("%d\n", mod);     }}