hduoj 5327 Olympiad (一维前缀和)

最新推荐文章于 2022-12-28 11:16:58 发布

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本文介绍了一种快速计算两个整数区间内所有美丽数数量的方法。美丽数定义为所有位数字均不相同的整数。通过预处理和使用一维前缀和技术，该算法能够高效地解决此类问题。

Problem 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 [a,b] (a≤b). Please be fast to get the gold medal!

Input

The first line of the input is a single integer T (T≤1000), indicating the number of testcases.
For each test case, there are two numbers a and b, as described in the statement. It is guaranteed that 1≤a≤b≤100000.

Output

For each testcase, print one line indicating the answer.

Sample Input

2
1 10
1 1000

Sample Output

10
738

思路：预处理+一维前缀和

#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 1e6 + 50;
int n,a,b;
int sum[MAXN];
bool vis[10];

bool isBeautiful(int x){
	memset(vis,0,sizeof(vis));
	while(x){
		int cur = x%10;
		if(!vis[cur]) vis[cur] = true;
		else return false;
		x /= 10;
	}
	return true;
}
void Init(){
	sum[0] = 0;
	for(int i = 1;i < MAXN; i++){
		sum[i] = isBeautiful(i) + sum[i-1];
	}
}
int main(){
	Init();
	cin>>n;
	while(n--){
		cin>>a>>b;
		cout<<sum[b] - sum[a-1]<<endl;
	} 
	return 0;
}