[模拟]uva10706 - Number Sequence

Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second

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…S_k. Each group S_k 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 <=25), 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                           Output for Sample Input

2

8

3

2

2

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Problem source: Iranian Contest

Special Thanks: Shahriar Manzoor, EPS.

思路：这道题暴力肯定超时，我想了很久，没想到什么方法，后来参照了他人的思路和代码。下面是题解

观察数字可以找到规律：

数字范围          数字位数范围    每个数字宽度   总共数字所占位数

1~9 ：              1~9.                     1                      45

10~99：           11~189                 2                      9000

100~999：       192~2889              3                     1386450

10000~9999:    2893~38889          4                     1881019000  

>=100000         38894 ~ *              5                      **

找出输入数字所在的数字位数范围，再进一步确定他具体所在的数字位数，再进行枚举。

比如说：输入 9045，由上表先可以确定它是在11~189数字位数范围内，再进一步可以确定他是在189（即：从1到99）这个位数范围里的，最后进行枚举，发现他恰好189这个位数范围内的数字99的个位上，因此输出 9 。

代码：

#include<iostream>#include<cmath>#include<cstdio>using namespace std;void solve(int num,int finish,int start,int len){    int j;    long long i,sum=0;    char ch[6];    for(i=0;i<finish;start+=len)    {        i=i+start;        if(num<=i)        {            i=i-start;            break;        }    }    int len1,cnum=num-i;    for(j=1;;j++)    {        len1=(int)log10(j)+1;        sum=sum+len1;        if(sum>=cnum) break;    }    sum-=len1;    sprintf(ch,"%d",j);    cout<<ch[cnum-sum-1]<<endl;    return;}int main(){    int n,m;    cin>>n;    while(n--)    {        cin>>m;        if(m<=45)            solve(m,45,1,1);        else if(m<=9000+45)        {            m=m-45;            solve(m,9000,11,2);        }        else if(m<=9000+45+1386450)        {            m=m-9045;            solve(m,1386450,192,3);        }        else if(m<=1386450+9045+188019000)        {            m=m-9045-1386450;            solve(m,188019000,2893,4);        }        else        {            m=m-188019000-9045-1386450;            solve(m,2147483647,38894,5);        }    }    return 0;}