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
#include <stdio.h> #include <cmath> #include <iostream> using namespace std; int len[32000] = {}; int main() { int t, n, i, sum = 0; for(i = 1; sum >= 0; i ++) { len[i] = len[i-1] + (int)log10(i) + 1; sum += len[i]; } cin >> t; while(t--) { cin >> n; //获得最后的sk i = 1; while(len[i] < n) n -= len[i++]; //获得sk中的最后一个数 i = 1; while((int)log10(i) + 1 < n) { n -= (int)log10(i) + 1; i++; } //i就是这个数 i /= pow(10, (int)log10(i) + 1 - n);//没有被打印出来的位数作为幂，没被打出来的是后半部分，直接除掉 i %= 10;//最后一位 cout << i << endl; } }