Algorithm-digital dp

• Note: The intervals in this article are all closed intervals

• LDU digital dp practice

• Explanation video
  $00:00--$ Digital$dp$ principle explanation
  $02:50--$ "The number of degrees" train of thought
  $21:40--$ "The number of degrees" code explanation
  $36:20--$ Explanation of theidea of ​​“number game”
  $48:00--$ "Number Game" code explanation
  $56:50--$ "Windy number" mentality explained
  $65:00--$ "Windy number" code explanation
  $83:30--$ "Number Game II" Train of Thought
  $94:20--$ "Number Game II" code explanation
  $104:10--$ "Don’t 62" train of thought
  $115:10--$ “Don’t 62” code explanation
  $127:40--$ "Hate 7 to break a wife" train of thought to explain
  $145:10--$ "Hate 7 not getting married" code explanation

• The problem solved by digital dp : find a given interval $[L,R]$ , in line with the functionf (x) \small f(x)The number of$f$$($$x$$)$ .
  Example:
  $f\left(x\right)$ : do not decrease the number-$x$ must satisfy the relationship that the digits from left to right are less than or equal to. Such as$123$，$446$
  $L=1$$,$ $R=19$
  Answer:$18$
  Explanation: Only$10$ is not a non-decreasing number, the rest are non-decreasing numbers, so the answer is$18$

• Digital dp skills :
  $1、$$dp(L,R)=dp(0,R)-dp(0,L-1)$
  For$[L,R]$ Interval problem, we generally convert it into two order$dp$ , ie find$[0,R]$ and$[0,L-1]$ Two paragraphs, and then subtract the results to get the[L, R] \small [L,R]we need$[L,R]$
  Example: Find the sum of the interval2, 2through the prefix sum
  $2,$ painting tree

• template

#include<vector>
#include<iostream>
using namespace std;

const int N = 100;

int f[N][N];
int u[N];

//根据题目要求来预处理f[i][j]
void init() {
    
    

}
//计算[0,n]符合f(x)的个数
int dp(int n) {
    
    
	//如果L可以取到0的话，特判-1
	if (n == -1) return 0;
	//特判0
	if (n == 0) return 0 / 1;

	//用于把每位的数字提取出来
	vector<int> vt;
	while (n) vt.push_back(n % 10), n /= 10;

	//res记录返回值，last保留前缀的信息
	int res = 0, last = 0;
	for (int i = vt.size() - 1; i >= 0; i--) {
    
    
		//第一种放法
		for (int j = 0; j < vt[i]; j++) {
    
    
			if (true) res += f[][];
		}
		//第二种放法就是vt[i]本身，特判vt[i]是否和法，不合法直接退出就好
		if (false) break;
	}

	//返回答案
	return res;
}
int main() {
    
    

	//预处理需要的数据
	init();

	int l, r;
	while (cin >> l >> r) {
    
    
		//技巧一
		cout << dp(r) - dp(l - 1) << endl;
	}
	return 0;
}