百练1019 Number Sequence

百练1019 Number Sequence题目出处：http://poj.grids.cn/practice/1019

题目描述：

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

就不翻译了。大致意思是说St=1234...t（如S11=1234567891011）。然后S1S2S3...Sk串成一个长的字符串，然后在这个大的字符串中找出第i个字符。

代码即思路。留意注释即可。

#include<stdio.h>
#include <math.h>
#define Length 99999            //由于读入的测试样例并不太大，因此该数值是合理的

int getlength(int num)
{
    return log10(num) + 1;
}

int getDigitInNum( int n, int num )  //求j的n位数字。如j:23578, n:2，则返回3
{
    int t = getlength(num) - n;
    while (t--)
        num /= 10;
    return num % 10;
}

int getDigitInS( int* s, int n ) 
{
    int k = 1;
    while (s[k] < n)
        k++;
    if (n == s[k])
        return k % 10;
    else
    {
        n = n - s[k-1];     
        return getDigitInNum(n, k); //要求的数字肯定的在j当中
    }
}

int getDigit( int* cas, int* s, int n ) 
{
    int k = 1;
    while (cas[k] < n)
        k++;
    if (n == cas[k])
        return k % 10;
    else                            //说明第n位在Sk当中
    {       
        n = n - cas[k-1];           //通过对n重新复制，转为求Sk中第n位数
        return getDigitInS(s, n);
    }
}

int main()
{
    int s[Length] = {0};
    int cas[Length] = {0};
    int i;
    
    //s[i]的值是串Si的长度，经检测无溢出
    for (i = 1; i < Length; ++i)
    {
        s[i] = s[i-1] + getlength(i);
    }
    //cas[i]的值是串S1S2S3...Si的长度
    for (i = 1; i < Length; ++i)
    {
        cas[i] = cas[i - 1] + s[i]; //cas[i]的值可能会有溢出
        if (cas[i] < 0)             //溢出时i的值为31268
            break;
    }
    i--;                            //溢出前的i，后面还会用到
    
    int testN;                      //测试样例个数
    scanf("%d", &testN);
    while (testN--)
    {
        int n, digit;   
        scanf("%d", &n);            //读入测试样例的值
        if (n > cas[i])             //溢出检测，用到之前的i值
            digit = getDigitInS(s, n - cas[i]);
        else
            digit = getDigit(cas, s, n);
        printf("%d\n", digit);
    }

    return 0;    
}