def amicable(num):
sum = 0
Amicable = {}
AMICABLE = []
AMI = {}
for number in range(1,num):
factsum = 0
for factors in range(1,int(number/2)+1):
if number % factors == 0:
factsum += factors
Amicable = factsum
for i in range(1,len(Amicable)):
for j in range(1,len(Amicable)):
if Amicable == j and Amicable == i and i != j:
AMICABLE.append(i)
AMICABLE.append(j)
AMICABLE = list(set(AMICABLE))
print("在%d以下的亲和数为：" %num)
print(AMICABLE)
for numbers in AMICABLE:
AMI = Amicable
sum += numbers
print("它们分别的因数和为：")
print(AMI)
print("它们的因数总和为：%d" %sum)
start_amicable = time.time()
amicable(10000)
time_amicable = time.time() - start_amicable
print("%f秒" %time_amicable)
在10000以下的亲和数为：
它们分别的因数和为：
{1184: 1210, 6368: 6232, 5564: 5020, 5020: 5564, 2924: 2620, 284: 220, 6232: 6368, 1210: 1184, 220: 284, 2620: 2924}
它们的因数总和为：31626
7.503518秒 def d(n):
div = []
i = 1
for i in range(1, n // 2 + 1):
if n % i == 0:
div.append(i)
#print (div)
sum = 0
for each in div:
sum+= each
#print(sum)
return sum
def is_amicable(n1, n2):
if (d(n1) == n2) and (d(n2) == n1) and (n1 != n2):
return True
return False
def calc(n):
am = []
while n :
a = d(n)
if is_amicable(a, n) and a != n:
am.append(a)
am.append(n)
n = n - 1
am = set(am)
print (am)
sum = 0
for each in am:
sum += each
print (sum) #include <stdio.h>
#include <time.h>
main()
{
int i, j, k, sum = 0, temp, x, y, z, a;
int begin, end;
begin = time(NULL);
for (i = 10; i < 10000; i++)
{
if (i == z) //当找到一个亲和数，它的相亲数直接跳过循环
{
sum += z;
goto Label;
}
k = i / 2;
temp = 0;
y = 0;
for (j = 2; j <= k; j++)//找真因子
{
if (i % j == 0)
{
temp += j;
}
}
temp++;
z = temp;
x = temp / 2;
for (j = 2; j <= x; j++)//找真因子
{
if (temp % j == 0)
{
y += j;
}
}
y++;
if (y == i && y != z)
{
sum += i;
a = z;
}
Label:continue;
}
end = time(NULL);
printf("%d\n", sum);
printf("time = %d", end - begin);
}
31626
time = 0
自己进行了小小的优化，应该还有很多提升空间，望大佬指教！ #计算10000以下所有相亲数之和
def amicable_number(n):
ami = []
sum = 0
for i in range(1, n):
if n % i == 0:
ami.append(i)
for each in ami:
sum += each
return sum
result = 0
for num in range(1, 10001):
a = amicable_number(num)
if amicable_number(a) == num and a != num:
result += num
print(result)
本帖最后由 guosl 于 2022-1-2 20:26 编辑
此题可以打表记录亲和数，用空间换时间。
/*
答案：31626
耗时：0.0072575秒
*/
#include <iostream>
using namespace std;
const int N = 10000;
char a;
int getFactorsSum(int n)//计算n的各因数之和
{
int d = (int)sqrt((double)n);
int s = 1;
for (int i = 2; i <= d; ++i)
{
if (n % i == 0)
{
int m = n / i;
if (m == i)
s += i;
else
s += i + m;
}
}
return s;
}
int main(void)
{
int nS = 0;
for (int i = 2; i <= N; ++i)
{
if (a == 0)
{
int k = getFactorsSum(i);//计算i的各因数之和
if (k != i)//各因数之和不能等于自身
{
int k1 = getFactorsSum(k);//计算k的各位因数之和
if (k1 == i) //i和k是互为亲和数
{
a = 1;
nS += i;
if (k <= N)
a = 1;
}
}
}
else
nS += i;
}
cout << nS << endl;
return 0;
}
import time as t
start = t.perf_counter()
def seeking_factor(a_num):
factors = []
if a_num >= 1:
factors.append(1)
else:
return factors
divisor = 2
upper_limit = a_num
while divisor < upper_limit:
if not a_num % divisor:
factors.append(divisor)
upper_limit = int(a_num / divisor)
factors.append(upper_limit)
divisor += 1
# factors.sort()
return factors
affinity_nums = []
num_bool = []
for i in range(10000):
num_bool.append(True)
for n in range(10000):
if num_bool:
d_n = sum(seeking_factor(n))
try:
num_bool
a = sum(seeking_factor(d_n))
if n == a and n != d_n:
affinity_nums.append(n)
affinity_nums.append(d_n)
num_bool = False
num_bool = False
else:
num_bool = False
except IndexError:
num_bool = False
print(affinity_nums)
print(sum(affinity_nums))
print("It costs %f s" % (t.perf_counter() - start))
结果：
31626
It costs 0.913826 s python
31626
时间：0.02099919319152832s
import time
tt=time.time()
n=10000
data=*n
for i in range(2,int(n**0.5+1)):
for j in range(i,n):
if i*j>n:
break
elif i==j:
data=data+i
else:
data=data+i+j
he=0
for i in range(1,n+1):
if data<=n and not data==i:
if data-1]==i:
he=he+i
print(i)
print(he)
print("时间：{}s".format(time.time()-tt))
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