uva 10706 Number Sequence（找规律）

uva 10706 Number Sequence

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

题目大意：有一个数字序列 1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011……

给出位数i，求该数字序列第i位为什么数字。

解题思路：具体看注释。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		int n;
		scanf("%d", &n);
		long long temp, move = 0, cnt = 0, dig = 0;
		while (move < n) {
			cnt++; //标记n在哪一组数据中
			temp = cnt;
			while (temp) {
				dig++; //dig为每一组数据的长度 1)1, 2)12, 3)123,...10)12345678910
				temp /= 10;
			}
			move += dig; //move为 1， 3， 6， 10， 15， 21， 28， 36... cnt组数据的长度
		}
		temp = cnt; //cnt组的最后一个数为cnt
		while (move != n) { //n为所求位数，move为cnt组数据总位数
			move--;
			temp /= 10; //当数据大于一位时， 通过此处使得数据和位数获得一致
			if (!temp) {
				cnt--;
				temp = cnt;
			}
		}
		printf("%lld\n", temp % 10);
	}
}