The fifteenth day of learning python---simple number theory

Modulo operation

(ab)mod m=(a mod m）（b mod m) mod m

1. Question statistics

a,b,n=map(int,input().split())
res=a*5+b*2
m=n//res*7
n=n%res
t=0
while n>0:
    if t<5:
        n-=a
        m+=1
    else:
        n-=b
        m+=1
print(m)

Two, fast power

b,p,k=map(int,input().split())
def quick_mi(b,p,k):
  n=p
  b%=k
  ans=1
  while n:
    if n&1==1:
      ans=(ans*b)%k
    b=(b*b)%k
    n>>=1
  return ans

print(quick_mi(b,p,k))

3. RSA decryption

First use the following code to solve p, q

import math
n=1001733993063167141
k=int(math.sqrt(n))
for i in range(2,k+1):
    if n%i==0:
        print(i,n//i)

891234941 1123984201

The second step is to convert the formula of de%((p-1)(q-1))

n=1001733993063167141
d=212353
p=891234941
q=1123984201
tmp=(p-1)*(q-1)
print(tmp)
for i in range(2,n+1):
    now=i*tmp+1
    if now%d==0:
        print(now//d)
        break

1001733991047948000
823816093931522017

The third step is to use the fast power code to solve x=c^e mod n

n=1001733993063167141
e=823816093931522017
c=20190324
def fastpow(a,b,mod):
    ans=1
    while b:
        if b&1:
            ans=ans*a%mod
        a=a*a%mod
        b>>=1
    return ans
print(fastpow(c,e,n))

579706994112328949

GCD

def gcd(a,b):
    if b==0:
        return a
    return gcd(b,a%b)

def gcd(a,b):
    return a if b==0 else gcd(b,a%b)

from math import gcd

LCM

Least common multiple
LCM=a*b/gcd(a,b)

def lcm(a,b):
	return a*b//gcd(a,b)

4. The number of walnuts (least common multiple)

This problem is to find the least common multiple of a, b, c

from math import *

def lcm(a,b):
    return a*b//gcd(a,b)

a,b,c=map(int,input().split())
ans=lcm(lcm(a,b),c)
print(ans)

Five, Hankson's fun questions

Passed 80%, the last one timed out, and did not find the python code of ac, if there is, please post it in the comment group!

import math
from math import gcd

def lcm(a,b):
    return a*b//gcd(a,b)

n=int(input())
for i in range(n):
    a0,a1,b0,b1=map(int,input().split())
    ans=0
    for x in range(1,int(math.sqrt(b1)+1)):
        if b1%x==0:
            if gcd(x,a0)==a1 and lcm(x,b0)==b1:
                ans+=1
            y=b1//x
            if x==y:continue
            if gcd(y,a0)==a1 and lcm(y,b0)==b1:
                ans+=1
    print(ans)

6. Find integers

I saw that other people's writing methods were looking for patterns before, which felt very complicated, but it felt like they were always looking for their least common multiple, and then the accumulated step size was very clever!

from math import *
mod=[0,0,1,2,1,4,5,4,1,2,9,0,5,10,11,14,9,0,11,18,9,11,11,15,17,9,23,20,25,16,29,27,25,11,17,4,29,22,37,23,9,1,11,11,33,29,15,5,41,46]
ans=2+mod[2]
k=2
for i in range(3,50):
    while True:
        if ans%i==mod[i]:
            k=lcm(k,int(i))
            break
        else:ans+=k#一直加他们的最小公倍数，一直到满足符合是其倍数
print(ans)

print(2022040920220409)

prime number

7. Stupid kid

import math
def check(u):
    if u<=1:
        return False
    for i in range(2,int(math.sqrt(u))+1):
        if u%i==0:
            return False
    return True

letter=[0]*26
s=input()
for i in range(len(s)):
    if "A"<=s[i]<"Z":
        letter[ord(s[i]) - ord('A')] += 1
    else:letter[ord(s[i])-ord('a')]+=1
letter.sort()
maxx=letter[-1]
for i in letter:
    if i==0:
        continue
    minn=i
    break
# print(maxx,minn)
if check(maxx-minn):
    print("Lucky Word")
    print(maxx-minn)
else:
    print("No Answer")
    print("0")

8. Prime numbers

N=10**6
primes=[]
bprime=[False]*N

def getPrimes(n):
    global primes
    global cnt
    bprime[0]=True
    bprime[1]=True
    for i in range(2,n+1):
        if not bprime[i]:
            primes.append(i)
            cnt+=1
            for j in range(i*2,n+1,i):
                bprime[j]=True

n=int(input())
cnt=0
getPrimes(n-1)

for p in primes:
    print(p,end=' ')
print()
print(cnt)

9. Decomposition of prime factors