POJ 1019 Number Sequence

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

输入

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)

输出

There should be one output line per test case containing the digit located in the position i.

样例输入

2

8

3

样例输出

2

2

题目跟之前的不大一样 之前是1-9的重复 这个是1-N的来

数学问题 转载的。。方法很简单 动笔写写就容易明白

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=31269;
unsigned a[maxn];
unsigned s[maxn];
void Set()
{
a[1]=1,s[1]=1;
for(int i=2;i<maxn;i++)
{
a[i]=a[i-1]+(int)log10((double)i)+1;
s[i]=s[i-1]+a[i];//表示第i组数字列的长度
}
}

int Compute(int n)
{
int i=1;
while(s[i]<n)
i++;
int pos=n-s[i-1];//答案在s[i-1]里的第pos个
int len=0;
for(i=1;len<pos;i++)
len+=(int)log10((double)i)+1;//找到对应len的位置
return (i-1)/(int)pow((double)10,len-pos)%10;
//此时的i-1便是具体的值 答案在i里  去除尾巴
}
//取出1234的2，那么多余的位数有2位：34。那么用1234 / 10^2，得到12，再对12取模10，就得到2

int main()
{
int t,n;
scanf("%d",&t);
Set();
while(t--)
{
scanf("%d",&n);
printf("%d\n",Compute(n));
}
return 0;
}