UVA 10706 Number Sequence

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

A single positive integer iis given. Write a program to find the digit located in the position iin the sequence of number groups S₁S₂…S_k. Each group S_kconsists 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

不擅长找规律的问题呢，如果可以的话还是直接预处理出来然后搞定。

#include<iostream>  #include<algorithm>#include<cmath>#include<cstdio>#include<string>#include<cstring>using namespace std;const int maxn = 31268;int f[maxn], i, j, m[maxn], p[maxn], t;int a[10] = { 1, 10, 100, 1000, 10000, 100000 };long long n, u;int main(){for (i = 0; i < maxn;i++)for (j = 1; j < 6; j++) if (i / a[j] == 0) {f[i] = j; break;}for (i = 1; i < maxn; i++) m[i] = m[i - 1] + f[i];for (i = 1; i < maxn; i++) p[i] = p[i - 1] + m[i];scanf("%d", &t);while (t--){cin >> n;for (i = maxn - 1; i >= 1; i--) if (n>p[i]) break;for (u = p[i], j = 1; j <= i + 1; j++){u += f[j];if (u>=n) break;}int k = f[j] - (n - u + f[j]);k = (j / a[k]) % 10;printf("%d\n", k);}return 0;}