POJ 1019 Number Sequence (the nth value of the loop increment sequence)

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35823		Accepted: 10340

Description

A single Positive An 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 integers numbers ranging from 1 to K, written one after another.
For example, the first digits of the sequence is 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 is one output line per test case containing the digit located in the position I.

Sample Input

283

Sample Output

22

To find out the value of the first position

Idea: You can divide the sequence into several groups such as 1  123   1234 ...  . Using s[]a[] Two arrays to record the length of each sequence, and the current length of each set of sequences, and then roughly find I in which group, and then in each

This value is found in the corresponding position in the small group.

 #include <iostream> #include <cstdio> #include < string.h> #include <string> #include <cmath> #include <queue> #define LL long longusing namespace std ; LL s[100000],a[100000];    ll SO (ll X,ll D) {while (d--) {x/=10; } return x%10;}    int main () {LL n,m,i,j,cla;    A[1]=s[1]=1;        for (i=2;i<=34000;i++)//First hit table {a[i]=a[i-1]+ (LL) log10 ((double) i) +1;    S[i]=s[i-1]+a[i];    } scanf ("%lld", &CLA);        while (cla--) {scanf ("%lld", &n);        i=0;        while (s[i]<n) i++;        m=n-s[i-1];//find subscript i=0 in n in all sequences; while (M>a[i]) i++;//i is the length of the N Group in printf ("%lld\n", So (i,a[i]-m));//<span style= "Font-family:consolas, ' Co Urier New ', Courier, mono, serif; line-height:18px; Background-color:rgb (248, 248, 248); " > For example, to remove 1234 of 2, the extra bits have 2 bits: 34. Then use 1234 / 100, get 12, then 12 modulo 10, get 2</span>} return 0;} 
 
 
 

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POJ 1019 Number Sequence (the nth value of the loop increment sequence)