解题报告之 HDU5327 Olympiad

最新推荐文章于 2021-11-20 12:20:36 发布

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最新推荐文章于 2021-11-20 12:20:36 发布

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分类专栏： ACM ACM_基础文章标签： ACM 水题 hdu5327 多校

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本文介绍了一种用于计算特定区间内美丽数数量的算法。美丽数是指所有位数均不相同的整数。通过预处理方式快速计算1到100000之间的美丽数累计数量，并利用这些数据解决给定区间内的美丽数计数问题。

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解题报告之 HDU5327 Olympiad

Description

You are one of the competitors of the Olympiad in numbers. The problem of this year relates to beatiful numbers. One integer is called beautiful if and only if all of its digitals are different (i.e. 12345 is beautiful, 11 is not beautiful and 100 is not beautiful). Every time you are asked to count how many beautiful numbers there are in the interval

. Please be fast to get the gold medal!

Input

The first line of the input is a single integer

, indicating the number of testcases.

For each test case, there are two numbers

and

, as described in the statement. It is guaranteed that

Output

For each testcase, print one line indicating the answer.

Sample Input

2
1 10
1 1000

Sample Output

10
738

题目大意：一个数被称为美丽数，如果它的每一位数都不一样。给出范围[L,R]，为这个范围内有多少美丽数。

分析：就是直接朴素的做法看看每个数是不是美丽数，打表保存1~100000分别有几个美丽数，直接输出sum[R]-sum[L-1]即可。

上代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;

const int MAXN = 1e5 + 10;
int num[MAXN];

void get()
{
	num[0] = 0;	
	for(int i = 1; i < MAXN; i++)
	{
		num[i] = num[i - 1];
		set<int> s;
		int tem = i;
		while(tem)
		{
			if(s.find( tem % 10 ) != s.end())
				break;
			s.insert( tem % 10 );
			tem /= 10;
		}
		if(tem == 0)
			num[i]++;
	}
}

int main()
{
	int n;
	cin >> n;
	get();
	while(n--)
	{
		int a, b;
		cin >> a >> b;
		cout << num[b] - num[a - 1] << endl;

	}
}