Accept: 189 Submit: 461
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem 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
3
1 10
1 100
1 1000
Sample Output
9
54
384
Source
“高教社杯”第三届福建省大学生程序设计竞赛某次群赛的一个很水的数位dp,但脑抽了,不知道想到哪里去了
#include <cstdio> #include <cstring> #include <iostream> using namespace std; typedef long long ll; int bits[20]; ll dp[20][15][2][2]; ll dfs(int pos, int pre, int iso, int doing) { if(pos == -1) return 1; int End = doing ? bits[pos] : 9; ll& ans = dp[pos][pre][iso][doing]; if(ans != -1) return ans; ans = 0; for(int i = 0; i <= End; ++i) { if(iso && i <= pre) ans += dfs(pos - 1, i, 0, doing && i == End); else if(!iso && i >= pre) ans += dfs(pos - 1, i, 1, doing && i == End); } return ans; } ll calc(ll x) { int ls = 0; memset(dp, -1, sizeof dp); while(x) { bits[ls++] = x % 10; x /= 10; } return dfs(ls - 1, 9, 1, 1); } int main() { ll n, m; int _; cin >> _; while(_ --) { cin >> n >> m; cout << calc(m) - calc(n - 1) << endl; } return 0; }