POJ 1019 Number Sequence

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 40208 Accepted: 11694

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

题意很容易理解

思路：将1 12 123 1234 。。。12345。。N分成n段分别记录他们的长度len【】和到这里的长度sum【】

注意  从10到99的数的长度为2  从100到999长度 为3//用到一个len【i】=len【i-1】+（int）log10（（double）i）+1；

对于每一个数，例如8  我们可以找到他在 第4段 找到他在该段的位置 为2。

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define N 40000long long len[N],sum[N];void init(){for(int i=1;i<N;i++){len[i]=len[i-1]+(int)log10((double)i)+1;sum[i]=sum[i-1]+len[i];}}int doit(int x){int i,j;while(sum[i]<x){i++;}x-=sum[i-1];//下边的pos和x是不一样的
int llen=0;for(i=1;llen<x;i++){llen+=(int)log10((double)i)+1;}int pos=llen-x;//这里的pos和x不同因为10—99为2；return (i-1)/(int)pow((double)10,pos)%10;}int main(){init();/*for(int i=1;i<=100;i++){printf("%d %d\n",len[i],sum[i]);}*/int ccase;scanf("%d",&ccase);while(ccase--){int n;scanf("%d",&n);int ans=doit(n);printf("%d\n",ans);}return 0;}