POJ:1019 Number Sequence

最新推荐文章于 2020-10-03 23:15:57 发布

原创最新推荐文章于 2020-10-03 23:15:57 发布 · 280 阅读

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给定一个正整数i，找出数字序列S1S2…Sk中第i个位置上的数字。每个Sk是1到k的连续正整数序列。文章提供了输入输出示例，并讨论了按行计算的方法，因为没有更优策略。

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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

思路：

一排排的算吧。。好像也确实没啥好的办法。当加的数字的位数大于等于两位的时候要用到log10(doublen)+1的公式来计算位数

#include<stdio.h>
#include<math.h>
typedef long long ll;
const int maxn = 65540;
ll a[maxn], b[maxn];

void reset()
{
	int i;
	a[1] = 1;
	b[1] = 1;
	for (i = 2; i <= 65537; ++i) //可以先大致估算一下要几排
	{
		a[i] = a[i - 1] + log10((double)i) + 1;
		b[i] = b[i - 1] + a[i];
	}
}

int main()
{
	reset();
	int t, i, m;
	ll pos, tmp;
	ll x;
	scanf("%d",&t);
	while (t--)
	{
		scanf("%lld",&x);
		for (i = 1; b[i] < x; ++i)
		{
			continue;
		}
		pos = x - b[i - 1];
		tmp = 0;
		for (m = 1; tmp < pos; m++)
		{
			tmp += (ll)log10((double)m) + 1;
		}
		m--;
		ll k = tmp - pos;
		m = m / (int)pow(10.0, k);
		m = m % 10;
		printf("%d\n",m);
	}
	return 0;
}

//65537