fzu2109--Mountain Number（数位dp）

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

看到题的第一反应是：数位dp。。那肯定不会。。。

感觉有点像记忆化搜索吧。cal(n)求小于等于n且满足条件的数。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 15;
int num
;
int dp

[2];    // dp[i][j][1] i位数 前一位为j

int dfs(int nowPos, int preNum, bool isPeak, bool isFirst, bool isMax)
{
if (nowPos == -1) return 1;

if (!isMax && dp[nowPos][preNum][isPeak]) return dp[nowPos][preNum][isPeak];

int ans = 0;
int limit = isMax ? num[nowPos] : 9;
for (int i = 0; i <= limit; ++i) {
// isFirst 是说前面都是0 也就是该位是第一位
if (isFirst && i == 0) ans += dfs(nowPos - 1, 9, false, true, false);
else if (isPeak && i >= preNum) ans += dfs(nowPos - 1, i, false, false, (isMax && i == limit));
else if (!isPeak && i <= preNum) ans += dfs(nowPos - 1, i, true, false, (isMax && i == limit));
}
if (!isMax) dp[nowPos][preNum][isPeak] = ans;

return ans;
}

int cal(int n)
{
int idx = 0;
while (n) {
num[idx++] = n % 10;
n /= 10;
}
return dfs(idx - 1, 9, false, true, true);
}

int main()
{
int t;
scanf("%d", &t);
while (t--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", cal(r) - cal(l - 1));
}
return 0;
}