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
3 1 10 1 100 1 1000
Sample Output
9 54 384
题意：
如果一个>0的整数x，满足a[2*i+1] >= a[2*i]和a[2*i+2]，则这个数为Mountain Number。
给出L， R，求区间[L, R]有多少个Mountain Number。
思路：
数位DP，判断当前是偶数位还是奇数位(从0开始)，
如果是偶数位，那么它要比前一个数的值小，
如果是奇数位，那么它要比前一个数的值大。
不明白为什么开long long 就答案错误
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
int a[30];
int dp[30][15][15];
int dfs(int pos,int pre,int sta,bool limit){if(pos<0){return 1;}if(!limit&&dp[pos][pre][sta]!=-1) return dp[pos][pre][sta];int up=limit?a[pos]:9;int tmp=0;for(int i=0;i<=up;i++){if(sta && i<=pre) tmp+=dfs(pos-1,i,0,limit && i==up);if(!sta && i>=pre) tmp+=dfs(pos-1,i,1,limit && i==up);}if(!limit) dp[pos][pre][sta]=tmp;return tmp;
}int solve(int x)
{memset(a,0,sizeof(a));int pos=0;while(x){a[pos++]=x%10;x/=10;}return dfs(pos-1,9,1,1);
}
int main()
{memset(dp,-1,sizeof(dp));int n,m;int T;cin>>T;while(T--){ scanf("%d%d",&n,&m);printf("%d\n",solve(m)-solve(n-1));}return 0;
}