X-Bi-peak Number-优快云博客

One integer number x is called "Mountain Number" if:

(1) x>0 and x is an integer;

(2) Assume x=a[0]a[1]...a[len-2]a[len-1](0≤a[i]≤9, a[0] is positive). Any a[2i+1] is larger or equal to a[2i] and a[2i+2](if exists).

For example, 111, 132, 893, 7 are "Mountain Number" while 123, 10, 76889 are not "Mountain Number".

Now you are given L and R, how many "Mountain Number" can be found between L and R (inclusive) ?

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only two integers L and R (1≤L≤R≤1,000,000,000).

Output

For each test case, output the number of "Mountain Number" between L and R in a single line.

Sample Input

Sample Output

9
54
384

从左至右记为a[0]、a[1]..........a[2*i]、a[2*i+1]......。相邻的奇数位小于等于偶数为的为特殊数。让求区间内特殊数的个数。奇偶还是第一次见到，感觉难点就在这，不过解决起来也是出乎意料的简单。然后就是细节的判断处理，一个模板套出来结果了。

代码如下：

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
int a[100];
LL dp[100][10][2];
//   pos  pre  sta
//         位      前一个   奇偶     前导0      边界
LL dfs(int pos, int pre, int sta, bool lead, bool limit)
{
	if (pos == -1)
	{
		return 1;
	}
	if (!limit && dp[pos][pre][sta] != -1)
		return dp[pos][pre][sta];
	int up = limit ? a[pos] : 9;
	LL temp = 0;
	for (int i = 0; i <= up; i++)
	{
		if (lead && i == 0) //(i == 0 && lead == 0) //有前导0 且 i = 0
			temp += dfs(pos - 1, 9, 0, lead && i == 0, limit && i == up);
		else if (sta == 1 && pre <= i)//奇数 且  pre小
			temp += dfs(pos - 1, i, 1 - sta, lead && i == 0, limit && i == up);
		else if (sta == 0 && pre >= i)//偶数 且  pre大
			temp += dfs(pos - 1, i, 1 - sta, lead && i == 0, limit && i == up);
		else
		{
			//cout << "***************" << endl;
		}
	}
	if (!limit)
		dp[pos][pre][sta] = temp;
	return temp;
}
LL solve(LL x)
{
	int pos = 0;
	while (x)
	{
		a[pos++] = x % 10;
		x /= 10;
	}
	return dfs(pos-1, 9, 0, 1, 1);
		//   pos  pre sta lead limit
}
int main()
{
	int t;
	LL le, ri;
	cin >> t;
	while (t--)
	{
		memset(dp, -1, sizeof(dp));

		cin >> le >> ri;
		cout << solve(ri) - solve(le - 1) << endl;
	}
	return 0;
}