Digital dp topic summary

digital dp

topic	knowledge points	difficulty
2376. Count Special Integers	digital dp template	difficulty
600. Non-negative integers without consecutive 1s	digital dp	difficulty
233. Number of 1s	Digital dp template questions	difficulty
902. Number combinations up to N	Digital dp template topic	difficulty
Number of 1's	Mask records the number of numbers x
The number of acwing 1081 degrees	Use base thinking + each bit is 1 or 0. The number of mask records 1
acwing 1082、1083 ，1085	What is the number of the first digit of the Mask record
acwing 1084	The mask records the sum of the previous numbers

Ideas for doing digital dp questions

First change the [a, b] interval to a single calculation from [0, a], [0, b]. then make a difference

Second, use the standard template code

Standard template code, according to the conditions defined in the topic, just modify the pre state variable.

Common state variables are

1. For the previous one, there is a divisive number between adjacent digits in this topic. Such as acwing's 1082, 1083, 1085
2. The sum of digits, refer to the case of acwing 1084
3. mask, numbers from 1-9 cannot be repeated. The mask record selects that number
4. $cn{t}_x$, the number of occurrences of x is used to count the number of occurrences. Number of reference numerals 1

    private static int dfs(int cur, int pre, boolean isNum, boolean isLimit) {
    
    
        if (cur == s.length) {
    
    
            return isNum ? 1 : 0;
        }
        // 记忆化
        int res = dp[cur][pre];
        if (isNum && !isLimit && res >= 0)
            return res;
        res = 0;
        // 不是数字，可以跳过
        if (!isNum) res += dfs(cur + 1, pre, false, false);
        // 枚举是数字的情况
        int up = isLimit ? s[cur] - '0' : 9;
        for (int i = isNum ? 0 : 1; i <= up; i++) {
    
    
            if (i == 4 || (pre == 6 && i == 2)) continue;
            res += dfs(cur + 1, i, true, isLimit && i == up);
        }
        // 记录
        if (!isLimit && isNum) dp[cur][pre] = res;
        return res;
    }