The Special Number

The Special Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1377    Accepted Submission(s): 380

Problem Description

In this problem, we assume the positive integer with the following properties are called ‘the special number’:
1) The special number is a non-negative integer without any leading zero.
2) The numbers in every digit of the special number is unique ,in decimal system.
Of course,it is easy to check whether one integer is a ‘special number’ or not, for instances, 1532 is the ‘special number’ and 101 is not. However, we just want to know the quantity of the special numbers that is less than N in this problem.

 

Input

The input will consist of a series of signed integers which are not bigger than 10,000,000. one integer per line. (You may assume that there are no more than 20000 test cases)

 

Output

For each case, output the quantity of the special numbers that is less than N in a single line.

 

Sample Input

10
12

 

Sample Output

9
10
 
/*
题意：
给你一个数n，要求你求出小于n的special number的个数。special number的定义：
1.没有前导0；
2.每位上的数字不一样，即101不是，111也不是

思路：
类似打表先求出1-10000000之间满足special number的数据存在a[]中
再用二分查询查找输入的n位于a[]的下标是多少，即答案。
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 10000005
int a[MAX] , j = 0 ;
void fun()
{
	int i ,b[10];
	for( i = 1; i <=10000000 ; i++) //枚举1到MAX之间的所有数据，判断是否满足special number的要求，满足则保存到a[]
	{
		int m = i ,flag = 0 ;
		memset(b,0,sizeof(b));
		while(m) //判断是否有位数相同
		{
			b[m%10] ++ ;
			if(b[m%10]>=2)
			{
				flag =1 ;
				break;
			}
			m /= 10 ;
		}
		if(!flag)
			a[j++] = i ;
	}
}

/*二分查找*/
int Binary_Search(int x)
{
	int low = 0 , high = j-1 ,mid ;
	while(low <= high)
	{
		mid = (low + high) / 2 ;
		if(a[mid] < x)
			low = mid + 1;
		else
			high = mid - 1 ;
	}
	return low ;
}
int main(void)
{
	fun();
	int m ;
	while(scanf("%d",&m) != EOF)
	{
		printf("%d\n",Binary_Search(m));
	}
	return 0;
}