带分数
def add_grades(n):
i = 1
time = 0
while i < n-1:
grades = n - i
for x in range(1,1000):
denominator = x
numerator = x * (n - i)
ls = []
for j in str(i):
ls.append(int(j))
for j in str(denominator):
ls.append(int(j))
for j in str(numerator):
ls.append(int(j))
if 0 not in ls and ls.count(1) == ls.count(2) == ls.count(3) == ls.count(4) == ls.count(5)\
== ls.count(6) == ls.count(7) == ls.count(8) == ls.count(9) == 1:
time += 1
#print(i,denominator,numerator)
i += 1
return time
n = int(input())
print(add_grades(n))李白打酒
num = 0
def drink(jiu,dian,hua):
if dian > 0:
drink(jiu*2,dian-1,hua)
if hua > 0:
drink(jiu-1,dian,hua-1)
if jiu==1 and dian==0 and hua==1:
global num
num += 1
return num
print(drink(2,5,10))
第39级台阶
ls = []
for x in range(1,40):
for y in range(1,20):
num = x + 2 * y
if num == 39 and (x+y) % 2 == 0:
#print(x,y,x+y)
ls.append((y,x+y))
def factor(m):
if m == 1:
return 1
elif m > 1:
return m * factor(m - 1)
def C(x,y):
return factor(y)//(factor(x) * factor(y-x))
time = 0
for i in ls:
time += C(i[0],i[1])
print(time)穿越雷区
迷宫
import collections
f=open("maze.txt","r")
s=f.readlines()
position={}
for y_wall in range(0,32):
position[(0,y_wall)]='1'
position[(51,y_wall)]='1'
for x_wall in range(0,52):
position[(x_wall,0)]='1'
position[(x_wall,31)]='1'
y=1
for y_position in s:
for xy_index in range(len(y_position)-1):
position[(xy_index+1,y)]=y_position[xy_index]
y=y+1
m_deque=collections.deque()
start=(1,1)
end=(50,30)
checked_position=[]
m_deque.append(start)
while m_deque:
to_check_position=m_deque.popleft()
if to_check_position not in checked_position:
if to_check_position==end:
print("ok!")
break
else:
checked_position.append(to_check_position)
up_position=(to_check_position[0],to_check_position[1]-1)
down_position=(to_check_position[0],to_check_position[1]+1)
left_position=(to_check_position[0]-1,to_check_position[1])
right_position=(to_check_position[0]+1,to_check_position[1])
if position.get(up_position)=="0" and up_position not in checked_position:
m_deque.append(up_position)
if position.get(down_position)=="0" and down_position not in checked_position:
m_deque.append(down_position)
if position.get(left_position)=="0" and left_position not in checked_position:
m_deque.append(left_position)
if position.get(right_position)=="0" and right_position not in checked_position:
m_deque.append(right_position)
all_line=checked_position+[end]
every_end=(end)
line=[]
line_dulr=[]
while every_end!=start:
up_xy=(every_end[0],every_end[1]-1)
down_xy=(every_end[0],every_end[1]+1)
left_xy=(every_end[0]-1,every_end[1])
right_xy=(every_end[0]+1,every_end[1])
if position.get(up_xy)=="0" and up_xy in all_line:
index=all_line.index(up_xy)
all_line=all_line[:index]
line.append(up_xy)
line_dulr.append("D")
every_end = up_xy
elif position.get(down_xy)=="0" and down_xy in all_line:
index=all_line.index(down_xy)
all_line=all_line[:index]
line.append(down_xy)
every_end=down_xy
line_dulr.append("U")
elif position.get(left_xy)=="0" and left_xy in all_line:
index=all_line.index(left_xy)
all_line=all_line[:index]
line.append(left_xy)
every_end=left_xy
line_dulr.append("R")
elif position.get(right_xy)=="0" and right_xy in all_line:
index=all_line.index(right_xy)
all_line=all_line[:index]
line.append(right_xy)
every_end=right_xy
line_dulr.append("L")
line_s=""
for i in line_dulr:
line_s=line_s+i
print(line_s[::-1])跳马
m,n = map(int,input().split(','))
total = 0
def jump(r,c,a,b):
global total
if r<0 or r>m or c<0 or c>n:
return
elif r+a== m and c+b== n:
total += 1
jump(r+a,c+b,1,2)
jump(r+a,c+b,1,-2)
jump(r+a,c+b,2,1)
jump(r+a,c+b,2,-1)
jump(0,0,0,0)
print(total)路径之谜
未名湖边的烦恼
import itertools
m,n = map(int,input().split(','))
ls = [1]*m + [-1]*n
def check(li):
stack = []
for i in li:
if i == 1:
stack.append(i)
else:
if stack == []:
return False
else:
stack.pop()
if sum(stack) >= 0:
return True
else:
return False
need = set()
time = 0
for i in itertools.permutations(ls):
need.add(i)
for m in need:
if check(m):
time += 1
print(time)大臣的旅费
cnt=0
node=0
def dfs(v,k): #进行dfs搜索
global cnt
global node
global vis
if k>cnt:
cnt=k
node=v
for i in range(len(E[v])):
if vis[E[v][i][0]]==False: #没访问过的点都进行一次dfs
vis[E[v][i][0]]=True #进行标记
dfs(E[v][i][0],k+E[v][i][1])
vis[E[v][i][0]]=False #回溯
n=int(input())
E=[[]for i in range(n+1)]
vis=[False for i in range(n+1)]
for i in range(n-1):
x,y,z=map(int,input().split())
E[x].append((y,z))
E[y].append((x,z)) #将各个点进行连接，用列表数组的方式进行
vis[1]=True
dfs(1,0) #找到距离1最远的点node
vis[1]=False
cnt=0
vis[node]=True
dfs(node,0) #从node出发，找到最远的点，就是所要求的最远距离
res=0
for i in range(1,cnt+1):
res+=i+10
print(res)
2n皇后问题
nums = eval(input())
n = 0
list_empty = []
while n < nums:
hang = [int(i) for i in input().split()]
list_empty.append(hang)
n +=1
def conflict(queen_list,new_queen,black=None):
num = len(queen_list)
try:
col = black[num]
except:
pass
try:
if len(black) != 0:
if new_queen == col:
return True
except:
pass
if list_empty[num][new_queen] == 0:
return True
for index,queen in enumerate(queen_list):
if abs(new_queen-queen) in (0,num-index):
return True
return False
def queen(num, queen_list=[]):
for new_queen in range(num):
if not conflict(queen_list, new_queen):
if len(queen_list) == num -1:
yield [new_queen]
else:
for result in queen(num,queen_list+[new_queen]):
yield [new_queen]+result
black_queen = list(queen(nums))
def queen_white(num,black,queen_list=[]):
for new_queen in range(num):
if not conflict(queen_list,new_queen,black):
if len(queen_list) == num -1:
yield [new_queen]
else:
for result in queen_white(num,black,queen_list+[new_queen]):
yield [new_queen]+result
end_num = []
sum = 0
for black in black_queen:
white_queens = list(queen_white(nums,black))
if len(white_queens)==0:
continue
end_num.append(len(white_queens))
for i in end_num:
sum +=i
print(sum)