poj 1019(组外偏移->组内偏移)-优快云博客

本文链接：https://blog.youkuaiyun.com/A775700879/article/details/16368359

本文详细阐述了如何通过编程找到数列特定位置的数字，涉及组合数学原理及算法实现，通过实例演示了解决步骤，适用于算法与数据结构的学习。

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Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31916		Accepted: 9075

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

2
8
3

Sample Output

2
2

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

简单的组合数学。思路是：先判断num是在第几组（题目中的Sk）,再计算出组内偏移（用num减去这个组之前的累计数字个数）。然后计算出那个组的组内偏移位置数字是多少。

需要细心一些。第一次样例没有过是因为sum[i] - num这个计算出来的实际上是组内的倒着数的偏移量，因此需要用belong[i]减去这个值才能得到正数偏移量。

每时每刻都需要知道每个值是从0计数还是从1计数。为了安全起见，本程序所有计数都统一从1开始。

提交记录：

1、Accepted！

代码：

/*Source Code

Problem: 1019		User: 775700879
Memory: 1192K		Time: 0MS
Language: G++		Result: Accepted
Source Code*/
#include 
    
    
     
     
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             #include 
             #define MAX_V 500 #define oo 0x3f3f3f3f #define UPPER 2147483647 using namespace std; long long belong[100000] = {0}; long long sum[100000] = {0}; long long strleng(int i) { long long result = 0; while (i != 0) { result++; i /= 10; } return result; } long long get(long long num, long long position) { int i = 0; while (num != 0) { if (i == position) return num % 10; num /= 10; i++; } } long long cal(long long rank, int group) { long long count = 0; long long i; for (i = 1; i <= group; i++) { count += strleng(i); if (count >= rank) { return get(i, count - rank); } } } int main() { int t; long long num; int i, j, k; sum[1] = belong[1] = 1; for (i = 2; ; i++) { belong[i] = belong[i-1] + strleng(i); sum[i] = sum[i-1] + belong[i]; if (sum[i] > UPPER) break; } cin >> t; while (t--) { cin >> num; for (i = 1; sum[i] < num && i < 100000; i++) { } cout << cal(belong[i] - (sum[i] - num), i) << endl; } return 0; }