Greatest English Letter in Upper and Lower Case

Given a string of English letters s, return the greatest English letter which occurs as both a lowercase and uppercase letter in s. The returned letter should be in uppercase. If no such letter exists, return an empty string.

An English letter b is greater than another letter a if b appears after a in the English alphabet.

Example 1:

Input: s = "lEeTcOdE"
Output: "E"
Explanation:
The letter 'E' is the only letter to appear in both lower and upper case.

找同一个字母大小写同时出现的，并且这个字母最靠后。

pii模拟

class Solution {
public:
    string greatestLetter(string s) {
        vector<pair<int, int>> m(26);
        for (auto &ch: s) {
            if (ch >= 'A' && ch <= 'Z') {
                m[ch - 'A'].first = 1;
            } else {
                m[ch - 'a'].second = 1;
            }
        }
        for (int i = 25; i >= 0; i--) {
            if (m[i].first == 1 && m[i].second == 1) {
                string t = "";
                t += 'A' + i;
                return t;
            }
        }
        return "";
    }
};

Sum of Numbers With Units Digit K

Given two integers num and k, consider a set of positive integers with the following properties:

• The units digit of each integer is k.
• The sum of the integers is num.

Return the minimum possible size of such a set, or -1 if no such set exists.

Note:

• The set can contain multiple instances of the same integer, and the sum of an empty set is considered 0.
• The units digit of a number is the rightmost digit of the number.

Example 1:

Input: num = 58, k = 9
Output: 2
Explanation:
One valid set is [9,49], as the sum is 58 and each integer has a units digit of 9.
Another valid set is [19,39].
It can be shown that 2 is the minimum possible size of a valid set.

给num反复的减去一个尾数为k的数，最后把num减没，否则就是-1；

特判一下0

然后贪心一下

然后最大次数

class Solution {
public:
    int minimumNumbers(int num, int k) {
        int ans = 0, times = 0;
        if (num == 0) { return 0; }
        while (num != 0) {
            if (num % 10 == k) {
                return ans + 1;
            } else {
                num = num - k;
                ans++;
            }
            if (++times > 10 || num < 0) { return -1; }
        }
        return -1;
    }
};

Longest Binary Subsequence Less Than or Equal to K

You are given a binary string s and a positive integer k.

Return the length of the longest subsequence of s that makes up a binary number less than or equal to k.

Note:

• The subsequence can contain leading zeroes.
• The empty string is considered to be equal to 0.
• A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Example 1:

Input: s = "1001010", k = 5
Output: 5
Explanation: The longest subsequence of s that makes up a binary number less than or equal to 5 is "00010", as this number is equal to 2 in decimal.
Note that "00100" and "00101" are also possible, which are equal to 4 and 5 in decimal, respectively.
The length of this subsequence is 5, so 5 is returned.

删除一些0 和1，然后让子串的十进制小于k

要保证出现更多的0和更多的1，也就是要看一下后面往前找的什么时候二进制大于k了，大于了就跳。否则就继续。

因为2^1000大于int128范围了，开个py吧

C++选手用py真的难受

class Solution:
    def longestSubsequence(self, s: str, k: int) -> int:
        pre=0;
        for  i in range(len(s)):
            if(s[i]=='0'):
                pre+=1
        i = len(s)-1 
        c = 1
        cnt=0
        while(i>=0):
            if(s[i]=='1'):
                cnt=cnt+c
            else:
                pre-=1
            if(cnt>k):
                break
            i-=1
            c=c*2
        return len(s)-i-1+pre
        

Selling Pieces of Wood

You are given two integers m and n that represent the height and width of a rectangular piece of wood. You are also given a 2D integer array prices, where prices[i] = [hi, wi, pricei] indicates you can sell a rectangular piece of wood of height hi and width wi for pricei dollars.

To cut a piece of wood, you must make a vertical or horizontal cut across the entire height or width of the piece to split it into two smaller pieces. After cutting a piece of wood into some number of smaller pieces, you can sell pieces according to prices. You may sell multiple pieces of the same shape, and you do not have to sell all the shapes. The grain of the wood makes a difference, so you cannot rotate a piece to swap its height and width.

Return the maximum money you can earn after cutting an m x n piece of wood.

Note that you can cut the piece of wood as many times as you want.

Example 1:

Input: m = 3, n = 5, prices = [[1,4,2],[2,2,7],[2,1,3]]
Output: 19
Explanation: The diagram above shows a possible scenario. It consists of:
- 2 pieces of wood shaped 2 x 2, selling for a price of 2 * 7 = 14.
- 1 piece of wood shaped 2 x 1, selling for a price of 1 * 3 = 3.
- 1 piece of wood shaped 1 x 4, selling for a price of 1 * 2 = 2.
This obtains a total of 14 + 3 + 2 = 19 money earned.
It can be shown that 19 is the maximum amount of money that can be earned.

裸题啊，dp枚举每个分割点就出了

老年人自信了，不应该太自信的去记录的。。。

class Solution {
public:
    long long dp[201][201];
    long long pre[201][201];

    long long sellingWood(int m, int n, vector<vector<int>> &prices) {
        for (auto &i: prices) {
            pre[i[0]][i[1]] = i[2];
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = pre[i][j];
                for (int k = 1; k < i; k++) {
                    dp[i][j] = max(dp[i][j], dp[k][j] + dp[i - k][j]);
                }
                for (int k = 1; k < j; k++) {
                    dp[i][j] = max(dp[i][j], dp[i][j - k] + dp[i][k]);
                }
            }
        }
        return dp[m][n];
    }
};