Mountain Number

最新推荐文章于 2020-05-26 11:02:39 发布

Rechard_chen

最新推荐文章于 2020-05-26 11:02:39 发布

阅读量715

点赞数

分类专栏：数位DP

本文链接：https://blog.youkuaiyun.com/Rechard_chen/article/details/44180331

版权

数位DP 专栏收录该内容

2 篇文章

订阅专栏

解决在给定范围内找出所有满足特定条件的山峰数字的问题，这些数字必须为正整数且满足中间数值大于两边的性质。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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

 
101 1001 1000 
  

Sample Output

954384

<span style="font-size:18px;"># include <cstdio>
# include <cstring>
# include <iostream>
using namespace std;

int dp[12][12][2];  //三维数组分别表示当前数位、前一位的数值、当前位属于的奇偶状态;
int doing,judge;   //doing是否达到临界，judge是否含有前导零;
int digit[12];  //储存数字的每一位：

int dfs(int pos,int pre,int pari,int judge,int doing)
{
    if(pos == -1)   return 1;
    if(!doing&& dp[pos][pre][pari] != -1)   return dp[pos][pre][pari];

    int ans = 0,end;
    end = doing ? digit[pos] : 9;

    for(int i = 0 ; i <= end;i++)
    {
        if(!(judge || i))   ans += dfs(pos-1,9,0,judge||i,doing &&( i == end));   //前一位和当前位都含前导零；
        else if(pari && pre <= i)  ans += dfs(pos-1,i,pari^1,judge||i,doing &&( i == end));
        else if(!pari&& pre >= i)  ans += dfs(pos-1,i,pari^1,judge||i,doing &&( i == end));
    }
    if(!doing)  dp[pos][pre][pari] = ans;
    return ans;
}

int cal(int x)<span style="white-space:pre">	</span>//取出数值每一位的数字;
{
    int cnt = 0;
    while(x > 0){
        digit[cnt++] = x % 10;
        x /= 10;
    }
    return dfs(cnt-1,9,0,0,1);  //当前数位，前一位数值，当前位奇偶，是否含有前导零，是否达到上界：
}

int main()
{
    int n;
    scanf("%d",&n);
    memset(dp,-1,sizeof(dp));
    while(n--){
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d\n",cal(r)-cal(l-1));
    }
    return 0;
}
</span>