POJ 1019 Number Sequence

首先说一下题意：给你一个数列，他是形如 1121231234123456123456712345678123……k。的数列，题意是让求出第几位数字是几。比如第一位是1，第80位是0。并不是说第八十个是几。

把数列进行分组：

1

12

123

1234

12345

……这样分组。

解题思路其实就是打表+推理，打表很简单就是算出前第几组时已经有多少个数字了。

然后就是分析，找到当前输入的n位于哪两个组之间，如果正好是这个组的末尾就输出k%10的结果，就是结果。如果不是的话，就找到位于两个数组的哪个位置，重点是判断位置，因为比如12是占两位的。所以用long10(i)经行递增。找到位置然后判断一下输出结果就OK了啊、、

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31675 Accepted: 9001

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

#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <cmath>#include <math.h>#define maxn 65537long long sum[65537], a[65537];using namespace std;void _printf(){    a[0] = sum[0] = 0;    int i;    for(i = 1; i < maxn; i++)        a[i] = a[i-1]+(int)log10(1.0*i) +1;    for(i = 1; i < maxn; i++)        sum[i] = sum[i-1]+a[i];}int main(){    int t, i;    long long n;    _printf();    cin >>t;    while(t--)    {        cin >>n;        int k;        for(i = 1; i < maxn; i++)        {            if(n <= sum[i])            {                k = i;                break;            }        }        if(sum[k] == n)        {            cout <<k%10<<endl;            continue;        }        int tt = 0;        int kk = n-sum[k-1];        for(i = 1; tt < kk; ++i)        {            tt += ((int)log10(1.0*i)+1);        }        cout <<(((i - 1)/(int)pow(10*1.0,1.0*(tt - kk)))%10)<<endl;    }    return 0;}