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笔试强训day05

news 2025/6/28 19:59:17

 
游游的you 
from math import *
from functools import *
from itertools import *
from bisect import *
def MII():return map(int, input().split())
def I():return input()
def II():return int(input())
def MSI():return map(str, input().split())
def LII():return list(int(x) for x in input().split())
def LSI():return list(s for s in input().split())
q = II()
for _ in range(q):a, b, c = MII()t1 = min(a, b, c)t2 = max(0, b - t1 - 1)print(t1 * 2 + t2)
 
 
 
腐烂的苹果 
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param grid int整型二维数组
# @return int整型
#
class Solution:def rotApple(self, grid: List[List[int]]) -> int:n,m = len(grid),len(grid[0])rot = {(i,j) for i in range(n) for j in range(m) if grid[i][j] == 2}fre = {(i,j) for i in range(n) for j in range(m) if grid[i][j] == 1}time = 0while fre:if not rot: #新鲜的苹果还存在，但是却没有苹果会腐烂，说明剩下的苹果都不会腐烂了，因此返回-1return -1rot = {(i+dx,j+dy) for i,j in rot for dx,dy in {(0,1),(0,-1),(1,0),(-1,0)} if (i+dx,j+dy) in fre} #在新鲜的苹果里找回变腐烂的苹果fre -= rottime += 1return time 
 
 
孩子们的游戏 
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param n int整型 
# @param m int整型 
# @return int整型
#
import sys
import functoolssys.setrecursionlimit(300000)class Solution:def LastRemaining_Solution(self , n: int, m: int) -> int:# write code here@functools.cachedef dfs(n,m)->int:if n == 1:return 0x = dfs(n-1,m)return (x+m)%nif n==0 or m==0:return -1return dfs(n,m)
 