Table of contents
1. Knowledge points
1. Sort
sorted () can be partially sorted by slicing, and the sort () function is intelligently used on the list
sorted implements custom sorting rules through custom keys, and needs to import functools
-1 means leave unchanged, 1 means swap order.
a sample question
import functools
def cmp(n1,n2): # 1表示交换，-1表示不变
if n1[1]!=n2[1]: # 降序
return -1 if n1[1]>n2[1] else 1
elif n1[2] != n2[2]: # 降序
return -1 if n1[2]>n2[2] else 1
else: # 升序
return 1 if n1[0]>n2[0] else -1
n = int(input())
scores =[]
for i in range(n):
score = list(map(int,input().split()))
score.append([i+1,sum(score)]+score) # [学号，总分，语数外]
scores=sorted(iterable,key=functools.cmp_to_key(cmp))
for i in range(5):
print(scores[i][0],scores[i][1])
2. Combination
Combination: combination in itertools
Permutation: permination in itertools
pigeonhole principle
Yang Hui Triangle
2. Brush questions
1. Two-way sorting
n,m=map(int,input().split())
a = [i for i in range(1,n+1)]
for i in range(m):# m次操作
p,q=map(int,input().split())
if p==0:
a=sorted(a[0:q:],reverse=True)+a[q::1] # 切片拼接 #下标从0开始，要减1
elif p==1:
a=a[:q-1:1]+sorted(a[q-1::1])
print(*a)
bid
n, m = map(int, input().split())
list1 = [i for i in range(1, n + 1)]
for j in range(m):
p, q = map(int, input().split())
if p == 0:
list2 = list1[0:q]
list2.sort()
list1[0:q] = list2[::-1]
else:
list2 = list1[q - 1::]
list2.sort()
list1[q - 1::] = list2
for i in list1:
print(i, end=' ')Timed out, over 60% data
numerical order
import os
import sys
# 请在此输入您的代码
n = int(input())
m = int(input())
x= [str(i) for i in range(1,n+1)]
import functools
def cmp(a,b): # -1不变 1 变
x1=sum(list(map(int,list(a)))) # 字符串拆分转列表求和
x2=sum(list(map(int,list(b))))
if x1!=x2:
return -1 if x1<x2 else 1 # 升序
else:
return -1 if a>b else 1 # 降序
x.sort(key=functools.cmp_to_key(cmp))
print(x[m-1])
bid
n = int(input())
m = int(input())
a = [i for i in range(1, n + 1)]
b = [0] * (n + 1)
for i in range(1, n + 1):
#求i的数位之和
num = i
while num != 0:
b[i] += num % 10
num //= 10
a.sort(key=lambda x: (b[x], x))
print(a[m - 1])swiss wheel
For those who have lost all in this round, the relative ranking between them will not change, and for those who have all won in this round, it has the same nature.
So every time for the end of each round, the sequence can be made orderly with only O(n) time complexity.
# 对于在该轮中全部输的人，他们之间的相对排名不会发生变化，对于在该轮中全部赢的人，也有同样的性质。
N, R, Q = map(int, input().split()) # 选手 ，比赛 ，名次
s = map(int, input().split()) # 初始分数
w = list(map(int, input().split())) # 实力值
i = range(2 * N)
data = sorted(map(lambda s, i: [s, i], s, i), key=lambda x: (-x[0], x[1])) # 构建[分数，索引] ，按分数从大到小排序
def merge(win, lose):
res = []
while win and lose:
if (win[0][0] == lose[0][0] and win[0][1] < lose[0][1]) or (
win[0][0] > lose[0][0]
): # 赢的总分高，或者总分一样但是索引更小
res.append(win.pop(0))
else:
res.append(lose.pop(0))
return res + win + lose
for _ in range(R):
win, lose = [], []
for i in range(0, 2 * N, 2): # 寻找胜利者失败者
if w[data[i][1]] > w[data[i + 1][1]]:
data[i][0] += 1
win.append(data[i])
lose.append(data[i + 1])
else:
data[i + 1][0] += 1
win.append(data[i + 1])
lose.append(data[i])
data = merge(win, lose)
print(data[Q - 1][1] + 1)Xiaolan eating candy problem
n = int(input())
a=list(map(int,input().split()))
if sum(a)-max(a) < max(a-1):
# 剩下的糖果数量小于隔板弄成的空间
# |1|2|3|4|5| 6板最少5空间
print('No')
else:
print("Yes")Sequence evaluation problem
As long as the last 4 digits, remember to take the remainder, otherwise the memory overflow is insufficient, even if there is no overflow, the calculation will be very slow
# maxn = 20190325
# a = [0] * maxn
# a[1], a[2], a[3] = 1, 1, 1
# for i in range(4, maxn):
# a[i] = (a[i-1] + a[i-2] + a[i-3]) % 10000
# print(a[20190324])
print(4659)Yang Hui Triangle
Fraudulent writing method (first print out and splice into a queue, then find the value through the list index) 20%
import os
import sys
# 骗分写法
n=int(input())
a=[[1],[1,1]]
for i in range(1,50): # 50-1+2行
b=[]
temp=0
for j in range(i): # 根据上一行i计算
temp=a[i][j]+a[i][j+1]
b.append(temp)
a.append([1]+b+[1])
# print(a)
b=[]
for i in range(51): #进行队列拼接
b=b+a[i]
print(b.index(n)+1) # 直接通过队列值找索引
Combination problem
Violent writing, over 20%
import os
import sys
# 请在此输入您的代码
def f(i,j): # 5 3 5*4*3/3*2*1
a,b=1,1
for z in range(i,i-j,-1): # 5,2
a*=z
for z in range(1,j+1):
b*=z
return a//b
t,k = map(int,input().split())
ans=0
for _ in range(t):
n,m=map(int,input().split())
for i in range(1,n+1):
for j in range(0,min(i,m)+1):
if f(i,j)%k==0:
print(i,j)
ans+=1
print(ans%(10**9+7))