标签：return Contest 313 self int while grid c2 Weekly

Weekly Contest 313

Problem A

Number of Common Factors

思路

数据范围小，直接暴力就完事了

代码

class Solution:
    def commonFactors(self, a: int, b: int) -> int:
        t = []
        for i in range(1,min(a,b)+1):
            if a%i==0 and b%i==0:
                t.append(i)
        return len(t)

Problem B

Maximum Sum of an Hourglass

思路

数据范围也挺小的，暴力就完事了

代码

class Solution:
    def maxSum(self, grid: List[List[int]]) -> int:
        ans = 0
        n = len(grid)
        m = len(grid[0])
        for i in range(0,n-2):
            for j in range(0,m-2):
                t = grid[i][j]+grid[i+2][j]
                t+=grid[i][j+1]+grid[i+1][j+1]+grid[i+2][j+1]
                t+=grid[i][j+2]+grid[i+2][j+2]
                ans = max(ans,t)
        return ans

Problem C

Minimize XOR

思路

转二进制算一下两个数各有多少1，构造结果，优先把1分配到num1的1的位置，多的1再从低到高分配，写的有点拖沓

代码

import numpy as np
class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        a,b,c = np.zeros(32,dtype=np.int),np.zeros(32,dtype=np.int),np.zeros(32,dtype=np.int)
        t = 0
        c1,c2 = 0,0
        t1,t2 = num1,num2
        while num1!=0:
            a[t] = num1%2
            if a[t]==1:
                c1+=1
            num1 = num1//2
            t+=1
        t = 0
        while num2!=0:
            b[t] = num2%2
            if b[t]==1:
                c2+=1
            num2 = num2//2
            t+=1
        if c1==c2:
            return t1
        l = 31
        while l>=0 and a[l] == 0:
            l-=1
        while l>=0 and c2!=0:
            if a[l]==1:
                c[l]=1
                c2-=1
            l-=1
        l = 0
        while c2!=0:
            if c[l]==0:
                c[l]=1
                c2-=1
            l+=1
        l = 31
        # print(c)
        while l>=0 and c[l]==0:
            l-=1
        t = 0
        while l>=0:
            # print(l)
            t = (t<<1)+c[l]
            # print(t)
            l-=1
        return t

Problem D

Maximum Deletions on a String

思路

数据范围小，暴力也能过,大致就是枚举长度做DFS

代码

class Solution:
    def deleteString(self, s: str) -> int:
        if (len(set(s)) == 1): return len(s)
        self.memo = {}
        return self.helper(s) + 1
        
    def helper(self, s):
        # print(s)
        if s in self.memo:
            return self.memo[s]
        total = 0
        for i in range(1, len(s) // 2 + 1):
            if s[:i] == s[i: i * 2]:
                total = max(total, self.helper(s[i:]) + 1)
        self.memo[s] = total
        return self.memo[s]

总结

出了三题，最后一题没整明白，没想到大家都是n^3我循环DP过不了 它DFS能过

标签：return,Contest,313,self,int,while,grid,c2,Weekly
From： https://www.cnblogs.com/baihualiaoluan/p/16770512.html