我编写了以下代码来计算所提供数字的最大素数因子 通过标准输入，但执行时间很长。如何减少程序的执行时间？
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool is_prime(unsigned long long num)
{
if (num == 0 || num == 1)
return false;
if (num == 2)
return true;
for (unsigned long long j = 2; j < num; j++)
if(num%j==0)
return false;
return true;
}
int main(void)
{
unsigned long long n,i;
printf("enter the number: ");
scanf("%llu",&n);
for(i=n-1;i>1;i--)
if((is_prime(i)) && n%i==0)
break;
printf("%llu\n",i);
return 0;
}
输入中的数字是600851475143。
答案 0 :(得分：3)
bool is_prime(unsigned long long num)
{
if (num < 2) return false; /* Reject border cases */
if (num == 2) return true; /* Accept the largest even prime */
if ((num & (~1)) == num) return false; /* Reject all other evens */
unsigned long long limit = sqrt(num) + 1; /* Limit the range of the search */
for (unsigned long long j = 3; j < limit; j += 2) /* Loop through odds only */
if(num % j==0)
return false; /* Reject factors */
return true; /* No factors, hence it is prime */
}
答案 1 :(得分：1)
保持相同的逻辑，你可以通过执行循环直到sqrt（num）来减少时间。 因为如果它不是素数，那么至少有一个除数小于sqrt（num）。
让我们拿num不是素数。所以对于某些1＆lt; a＆lt; num，a * b = num。
所以a或b应小于sqrt（num）。
因此，您的is_prime代码应更改为如下
bool is_prime(unsigned long long num)
{
if (num == 0 || num == 1)
return false;
if (num == 2)
return true;
long long root_num = sqrt(num);
for (unsigned long long j = 2; j < root_num; j++)
if(num%j==0)
return false;
return true;
}
现在，如果你想改变你的逻辑，那么有一种称为筛选方法的算法可以找到小于给定数量的所有素数。看看Sieve Algorithm
答案 2 :(得分：1)
main（）循环中的if语句应首先检查n％i == 0，以便不检查非因子的完整性。在一个现代英特尔CPU上运行的简单测试案例中，仅此一项就可以使程序从1到2000的所有数字快200倍。下面的mod只是对main（）（我把它分解成一个返回最大素数因子的函数），显着提高了速度。我也认为数字本身是一个可能的因素（原文中没有这样做）。
unsigned long long largest_prime_factor(unsigned long long n)
{
unsigned long long i,sqrtn,maxprime;
if (is_prime(n))
return(n);
sqrtn=(unsigned long long)sqrt((double)n);
maxprime=1;
for (i=2;i<=sqrtn;i++)
{
if (n%i==0)
{
unsigned long long f;
f=n/i;
if (is_prime(f))
return(f);
if (is_prime(i))
maxprime=i;
}
}
return(maxprime);
}
答案 3 :(得分：1)
玩得开心...所有素因子的一个函数，一个用于最大素因子：
#include <stdio.h>
#include <stdlib.h>
typedef unsigned long long u64_t;
static size_t prime_factors(u64_t value, u64_t *array)
{
size_t n_factors = 0;
u64_t n = value;
u64_t i;
while (n % 2 == 0)
{
n /= 2;
//printf("Factor: %llu (n = %llu)\n", 2ULL, n);
array[n_factors++] = 2;
}
while (n % 3 == 0)
{
n /= 3;
//printf("Factor: %llu (n = %llu)\n", 3ULL, n);
array[n_factors++] = 3;
}
for (i = 6; (i - 1) * (i - 1) <= n; i += 6)
{
while (n % (i-1) == 0)
{
n /= (i-1);
//printf("Factor: %llu (n = %llu)\n", (i-1), n);
array[n_factors++] = (i-1);
}
while (n % (i+1) == 0)
{
n /= (i+1);
//printf("Factor: %llu (n = %llu)\n", (i+1), n);
array[n_factors++] = (i+1);
}
}
if (n != 1)
array[n_factors++] = n;
return n_factors;
}
static u64_t max_prime_factor(u64_t value)
{
u64_t n = value;
u64_t i;
u64_t max = n;
while (n % 2 == 0)
{
n /= 2;
//printf("Factor: %llu (n = %llu)\n", 2ULL, n);
max = 2;
}
while (n % 3 == 0)
{
n /= 3;
//printf("Factor: %llu (n = %llu)\n", 3ULL, n);
max = 3;
}
for (i = 6; (i - 1) * (i - 1) <= n; i += 6)
{
while (n % (i-1) == 0)
{
n /= (i-1);
//printf("Factor: %llu (n = %llu)\n", (i-1), n);
max = (i-1);
}
while (n % (i+1) == 0)
{
n /= (i+1);
//printf("Factor: %llu (n = %llu)\n", (i+1), n);
max = (i+1);
}
}
//printf("Max Factor = %llu\n", (n == 1) ? max : n);
return (n == 1) ? max : n;
}
static void print_factors(u64_t value, size_t n_factors, u64_t *factors)
{
printf("%20llu:", value);
int len = 21;
for (size_t i = 0; i < n_factors; i++)
{
if (len == 0)
len = printf("%.21s", "");
len += printf(" %llu", factors[i]);
if (len >= 60)
{
putchar('\n');
len = 0;
}
}
if (len > 0)
putchar('\n');
}
void builtin_tests(void)
{
u64_t tests[] =
{
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 25, 49, 63, 69, 71, 73, 825, 829, 1000, 6857, 73112,
731122, 7311229, 73112293, 731122935, 7311229357, 73112293571,
600851475143, 731122935711, 7311229357111,
18446744073709551610ULL, 18446744073709551611ULL,
18446744073709551612ULL, 18446744073709551613ULL,
18446744073709551614ULL, 18446744073709551615ULL,
};
u64_t factors[64];
for (size_t i = 0; i < sizeof(tests)/sizeof(tests[0]); i++)
{
u64_t max = max_prime_factor(tests[i]);
printf("%20llu -> Max Prime Factor = %llu\n", tests[i], max);
size_t n_factors = prime_factors(tests[i], factors);
print_factors(tests[i], n_factors, factors);
}
}
int main(int argc, char **argv)
{
if (argc == 1)
builtin_tests();
else
{
for (int i = 1; i < argc; i++)
{
u64_t factors[64];
u64_t value = strtoull(argv[i], 0, 0);
u64_t max = max_prime_factor(value);
printf("%20llu -> Max Prime Factor = %llu\n", value, max);
size_t n_factors = prime_factors(value, factors);
print_factors(value, n_factors, factors);
}
}
return 0;
}
1 -> Max Prime Factor = 1
1:
2 -> Max Prime Factor = 2
2: 2
3 -> Max Prime Factor = 3
3: 3
4 -> Max Prime Factor = 2
4: 2 2
5 -> Max Prime Factor = 5
5: 5
6 -> Max Prime Factor = 3
6: 2 3
7 -> Max Prime Factor = 7
7: 7
8 -> Max Prime Factor = 2
8: 2 2 2
9 -> Max Prime Factor = 3
9: 3 3
10 -> Max Prime Factor = 5
10: 2 5
11 -> Max Prime Factor = 11
11: 11
12 -> Max Prime Factor = 3
12: 2 2 3
13 -> Max Prime Factor = 13
13: 13
14 -> Max Prime Factor = 7
14: 2 7
15 -> Max Prime Factor = 5
15: 3 5
16 -> Max Prime Factor = 2
16: 2 2 2 2
17 -> Max Prime Factor = 17
17: 17
18 -> Max Prime Factor = 3
18: 2 3 3
19 -> Max Prime Factor = 19
19: 19
20 -> Max Prime Factor = 5
20: 2 2 5
25 -> Max Prime Factor = 5
25: 5 5
49 -> Max Prime Factor = 7
49: 7 7
63 -> Max Prime Factor = 7
63: 3 3 7
69 -> Max Prime Factor = 23
69: 3 23
71 -> Max Prime Factor = 71
71: 71
73 -> Max Prime Factor = 73
73: 73
825 -> Max Prime Factor = 11
825: 3 5 5 11
829 -> Max Prime Factor = 829
829: 829
1000 -> Max Prime Factor = 5
1000: 2 2 2 5 5 5
6857 -> Max Prime Factor = 6857
6857: 6857
73112 -> Max Prime Factor = 37
73112: 2 2 2 13 19 37
731122 -> Max Prime Factor = 52223
731122: 2 7 52223
7311229 -> Max Prime Factor = 24953
7311229: 293 24953
73112293 -> Max Prime Factor = 880871
73112293: 83 880871
731122935 -> Max Prime Factor = 89107
731122935: 3 5 547 89107
7311229357 -> Max Prime Factor = 7311229357
7311229357: 7311229357
73112293571 -> Max Prime Factor = 8058227
73112293571: 43 211 8058227
600851475143 -> Max Prime Factor = 6857
600851475143: 71 839 1471 6857
731122935711 -> Max Prime Factor = 4973625413
731122935711: 3 7 7 4973625413
7311229357111 -> Max Prime Factor = 12939061
7311229357111: 547 1033 12939061
18446744073709551610 -> Max Prime Factor = 1504703107
18446744073709551610: 2 5 23 53301701 1504703107
18446744073709551611 -> Max Prime Factor = 287630261
18446744073709551611: 11 59 98818999 287630261
18446744073709551612 -> Max Prime Factor = 2147483647
18446744073709551612: 2 2 3 715827883 2147483647
18446744073709551613 -> Max Prime Factor = 364870227143809
18446744073709551613: 13 3889 364870227143809
18446744073709551614 -> Max Prime Factor = 649657
18446744073709551614: 2 7 7 73 127 337 92737 649657
18446744073709551615 -> Max Prime Factor = 6700417
18446744073709551615: 3 5 17 257 641 65537 6700417
答案 4 :(得分：0)
最好从数字的平方根开始循环到3， 再次，你可以跳过偶数。
这样你就可以在很短的时间内得到答案。