Mountain Number（数位dp）

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本文介绍了一种使用数位DP算法解决“山数”问题的方法。山数是一种特殊的整数，其奇数位置的数字必须大于等于相邻的两个数字。文章详细解释了如何通过递归函数计算在给定范围内山数的数量，并提供了完整的C++代码实现。

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https://cn.vjudge.net/problem/FZU-2109

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

x=a[0]a[1]...a[len-2]a[len-1](0≤a[i]≤9, a[0] is positive)

下标为奇数的要大于等于两边

数位dp，注意ll用I64d

#include<cstdio>
#include<cstring>
#pragma GCC optimize(3)
#define max(a,b) a>b?a:b
using namespace std;
typedef long long ll;
ll dp[15][2][10];
int a[15];
ll dfs(int pos,int sta,int pre,bool lead,bool limit){
	if(pos==0) return 1;
	if(!lead&&!limit&&dp[pos][sta][pre]!=-1) return dp[pos][sta][pre];
	int up=limit?a[pos]:9;
	ll ans=0;
	for(int i=0;i<=up;i++){
		if(lead&&i==0) ans+=dfs(pos-1,0,9,true,limit&&i==up);
		else if(sta==0&&i<=pre) ans+=dfs(pos-1,1,i,false,limit&&i==up);
		else if(sta==1&&i>=pre) ans+=dfs(pos-1,0,i,false,limit&&i==up);
	}
	if(!lead&&!limit) dp[pos][sta][pre]=ans;
	return ans;
}
ll solve(ll x){
	int pos=0;
	while(x){
		a[++pos]=x%10;
		x/=10;
	}
	return dfs(pos,0,9,true,true);
}
int main(){
    memset(dp,-1,sizeof(dp));
    int T;
    scanf("%d",&T);
    while(T--){
    	ll L,R;
    	scanf("%I64d%I64d",&L,&R);
    	printf("%I64d\n",solve(R)-solve(L-1));
	}
	return 0;
}