算法-数论
1、最大公约数
def gcd(a,b):
if b == 0:
return a
return gcd(b, a%b) # a和b的最大公约数等于b与a mod b 的最大公约数
def gcd(a,b):
while b != 0:
cur = a
a = b
b = cur%b
pass
return a
例题：
1、蓝桥杯2019年第十届省赛真题-等差数列 - C语言网 (dotcpp.com)
import math
n=int(input())
arr=list(map(int,input().split()))
if arr[0] > arr[1]:
max_ = arr[0]
min_ = arr[1]
else:
max_ = arr[1]
min_ = arr[0]
sub = max_ -min_
for i in range(2, n):
sub = math.gcd(sub, abs(arr[i]-arr[i-1]))
min_ = min(min_, arr[i])
max_ = max(max_, arr[i])
if max_ == min_ : print(n)
else:
print((max_ - min_) // sub + 1)
2、1223. 最大比例 - AcWing题库
辗转相减法
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
def gcd(a,b):
if b == 0:
return a
return gcd(b, a%b)
top, bot = 0,0
def gcd_sub(a, b):
if a < b:
a,b = b,a
if b==1:
return a
return gcd_sub(b, a//b)
i = 1
while i < n:
if arr[i] != arr[0]:
gcd_ = gcd(arr[i], arr[0])
top = arr[i]//gcd_
bot = arr[0]//gcd_
break
i += 1
if i == n:
print(1)
else:
for i in range(i, n):
gcd_ = gcd(arr[i], arr[0])
a = arr[i]//gcd_
b = arr[0]//gcd_
top = gcd_sub(a, top)
bot = gcd_sub(b, bot)
print(f'{top}/{bot}')
1.1 扩展欧几里得定律
def ext_euclid(a, b):
if b == 0:
return 1, 0, a
else:
x, y, q = ext_euclid(b, a % b)
# q = gcd(a, b) = gcd(b, a%b)
x, y = y, (x - (a // b) * y)
return x, y, q
扩展欧几里得算法求系数x，y_哔哩哔哩_bilibili
例题：1299. 五指山 - AcWing题库
def olai(a, b):
if b == 0:
return 1,0,a
x, y, gcd = olai(b, a%b)
x, y = y, x-(a//b)*y
return x, y, gcd
def funt(n, d, x, y):
x1, y1, gcd = olai(n, d)
if (y-x)%gcd != 0:
print('Impossible')
else:
d = y1*(y-x)//gcd
n //= gcd # ax+by = n; x = x+k*(b/gcd(a,b)); y = y+k*(a/gcd(a,b))
print((d%n+n)%n)
for _ in range(int(input())):
n, d, x, y = map(int, input().split())
funt(n, d, x, y)
2、最小公倍数
def funt(a,b):
return a*b//gcd(a,b)
3、质数筛
def getPrimes(n):
is_primes = [True]*(n+1) # 初始化一个布尔数组，表示从2到n的所有数都是质数
primes = [] # 存储质数
for i in range(2, int(n**(1/2))+1):
if is_primes[i]:
primes.append(i)
# 将当前质数的所有倍数标记为非质数
for j in range(i*i, n+1, i):
is_primes[j] = False
# 后面的质数的倍数一定会超
for i in range(int(n**(1/2))+1, n+1):
if is_primes[i]:
primes.append(i)
return primes
4、区间质数筛
import math
def simple_sieve(limit):
primes = []
sieve = [True] * (limit + 1)
# 0和1不是质数，所以标记为False
sieve[0] = sieve[1] = False
# 从2到根号limit遍历数字
for num in range(2, int(math.sqrt(limit)) + 1):
if sieve[num]:
primes.append(num)
for multiple in range(num * num, limit + 1, num):
sieve[multiple] = False
for num in range(int(math.sqrt(limit)) + 1, limit + 1):
if sieve[num]:
primes.append(num)
return primes
def segmented_sieve(start, end):
# 计算质数的上限
limit = int(math.sqrt(end)) + 1
primes = simple_sieve(limit)
primes_count = len(primes)
# 创建一个布尔数组，用于标记区间内的数字是否为质数，初始化为True
sieve = [True] * (end - start + 1)
# 对于每一个小于等于根号end的质数
for i in range(primes_count):
# 计算刚好大于等于start的primes[i]的数
base = max(primes[i]*primes[i], ((start + primes[i] - 1) // primes[i]) * primes[i])
# 将当前质数的倍数标记为非质数
for j in range(base, end + 1, primes[i]):
sieve[j - start] = False
# 生成区间内的质数列表
segmented_primes = [start + i for i in range(end - start + 1) if sieve[i]]
return segmented_primes
start = 10
end = 50
segmented_sieve(start, end)
5、欧拉函数
def funt(n):
count = n
p = 2
while p*p <= n:
# 找到质因子
if n % p == 0:
while n % p == 0:
n //= p
count -= count//p # n*(1-p)
p += 1
if n > 1:
count -= count//n
return count
funt(20)
欧拉函数公式证明_哔哩哔哩_bilibili
6、计算质因子个数
def funt(n):
count = 0
p = 2
while p * p <= n:
if n % p == 0:
count += 1
while n % p == 0:
n //= p
p += 1
if n > 1:
count += 1
return count
funt(12)
7、计算所有约数个数
约数，又称因数。整数a除以整数b(b≠0) 除得的商正好是整数而没有余数，就说a能被b整除，或b能整除a。a称为b的倍数，b称为a的约数。
def funt(n):
count = 1
p = 2
while p*p <= n:
cnt = 0
while n%p == 0:
cnt += 1
n //= p
count *= (cnt+1)
if n>1:
count *= 2
return count
funt(12)
8、计算所有约数和
def funt(n):
cnt = 0
p = 1
while p*p <= n:
# 一次算两个约数
if n%p == 0:
cnt += p
# 平方就只用算一次
if p*p != n:
cnt += n//p
p += 1
return cnt
funt(12)
例题
1295. X的因子链 - AcWing题库
# 方法一 动规
def funt(n):
primes = [n]
p = 2
while p*p <= n:
if n%p == 0:
primes.append(p)
if p*p != n:
primes.append(n//p)
p += 1
if n>1 and n != primes[0]:
primes.append(n)
primes.sort()
dp = [[1]*len(primes) for _ in range(2)]
for i in range(1, len(primes)):
max_ = cnt = 0
for j in range(i):
if primes[i] % primes[j] == 0:
if dp[0][j] > max_:
cnt = dp[1][j]
max_ = dp[0][j]
elif dp[0][j] == max_:
cnt += dp[1][j]
dp[0][i],dp[1][i] = max_+1, cnt if cnt else 1
print(dp[0][-1], dp[1][-1])
while True:
try:
n = int(input())
funt(n)
except EOFError:
break
# 数学！！！ https://www.acwing.com/solution/content/97535/ 具体请看题解，我实在不知道怎么表达o(´^｀)o
def A(a, b):
cnt = 1
while b > 0:
cnt *= a
a -= 1
b -= 1
return cnt
def funt(n):
cnt = []
p = 2
while p*p <= n:
if n%p == 0:
cnt_ = 0
while n%p == 0:
cnt_ += 1
n //= p
cnt.append(cnt_)
p += 1
if n>1:
cnt.append(1)
sum_ = sum(cnt)
a = 1
for i in cnt:
if i > 1:
a *= A(i,i)
times = A(sum_, sum_)//a
print(sum_, times)
while True:
try:
n = int(input())
funt(n)
except EOFError:
break
9、快速幂
# n为负数则最终结果返回 1/pow(x,-n)
def pow(x, n):
if n < 2:
return x**n
ret = pow(x, n//2)
if n%2:
return ret*ret*x
return ret*ret
pow(2, 10)
## 9、快速幂
```Python
# n为负数则最终结果返回 1/pow(x,-n)
def pow(x, n):
if n < 2:
return x**n
ret = pow(x, n//2)
if n%2:
return ret*ret*x
return ret*ret
pow(2, 10)
