Created
September 18, 2019 02:10
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| """ | |
| 求不大于n的所有数质数分解后的质数总个数 | |
| 快手笔试第3题 | |
| """ | |
| from random import randint | |
| def quickMulMod(a,b,m): | |
| ret = 0 | |
| while b: | |
| if b&1: | |
| ret = (a+ret)%m | |
| b//=2 | |
| a = (a+a)%m | |
| return ret | |
| def quickPowMod(a,b,m): | |
| ret =1 | |
| while b: | |
| if b&1: | |
| ret =quickMulMod(ret,a,m) | |
| b//=2 | |
| a = quickMulMod(a,a,m) | |
| return ret | |
| def isPrime(n,t=5): | |
| t = min(n-3,t) | |
| if n<2: | |
| return | |
| if n==2: return True | |
| d = n-1 | |
| r = 0 | |
| while d%2==0: | |
| r+=1 | |
| d//=2 | |
| tested=set() | |
| for i in range(t): | |
| a = randint(2,n-2) | |
| while a in tested: | |
| a = randint(2,n-2) | |
| tested.add(a) | |
| x= quickPowMod(a,d,n) | |
| if x==1 or x==n-1: continue #success, | |
| for j in range(r-1): | |
| x= quickMulMod(x,x,n) | |
| if x==n-1:break | |
| else: | |
| return False | |
| return True | |
| def gcd(a,b): | |
| while b!=0: | |
| a,b=b,a%b | |
| return a | |
| cache={1:0} | |
| def factor(n): | |
| if n in cache: | |
| return cache[n] | |
| if n==1: | |
| return 0 | |
| if isPrime(n): | |
| return 1 | |
| fact=1 | |
| cycle_size=2 | |
| x = x_fixed = 2 | |
| c = randint(1,n) | |
| while fact==1: | |
| for i in range(cycle_size): | |
| if fact>1:break | |
| x=(x*x+c)%n | |
| if x==x_fixed: | |
| c = randint(1,n) | |
| continue | |
| fact = gcd(x-x_fixed,n) | |
| cycle_size *=2 | |
| x_fixed = x | |
| cache[n]=factor(fact)+factor(n//fact) | |
| return cache[n] | |
| s=0 | |
| for i in range(2,1+int(input())): | |
| s+=factor(i) | |
| print(s) |
Author
m2kar
commented
Sep 18, 2019


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