blocks|key|3962589|text|在内部循环中，如果确实存在一个不是i因子的数字，则设置为b+=+i。如果不存在作为i因子的数字，则需要设置b+=+i。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3962590|(我所说的“数字”，当然是指“2和sqrt(i)之间的整数”)|3962591|entityMap^0|H|1|S|5|15|1|1H|5|0|F|1|H|7|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$]]

In your inner loop, you're setting <code>b = i</code> if there does exist a number that isn't a factor of <code>i</code>. You need to set <code>b = i</code> if there doesn't exist a number that is a factor of <code>i</code>.

(by "number", I mean "an integer between <code>2</code> and <code>sqrt(i)</code>" of course)

blocks|key|683724|text|要找到素因式分解，通常会找到2和sqrt(N)之间的所有因子。您可以将组合除以其中的每一个，以获得其余的因子。然后递归找出每一个的质因数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|683725|完成后，您将拥有一个包含所有主要因素的列表。获取列表中最大的项应该是相当简单的。|683726|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

To find the prime factorization, you'd normally find all the factors between 2 and sqrt(N). You'd divide the composite by each of those to obtain the rest of the factors. Then recurse to find the prime factors of each of those.

When you're done, you'll have a list of all the prime factors. Getting the largest item in the list should be fairly trivial.

blocks|key|2596012|text|因式分解是一个经典的数论问题。你可以在数论教科书中找到许多因式分解算法。其中一些可以在维基http://en.wikipedia.org/wiki/Integer_factorization上找到|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2596013|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Integer_factorization^0|19|1E|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Factorization is a classic number theory problem. You can find many factorization algorithms in number theory textbooks. Some of them are available on wiki <a href="http://en.wikipedia.org/wiki/Integer_factorization" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Integer_factorization</a>

blocks|key|2604142|text|嘿，谢谢你的输入，朋友们，我想出了一个办法，现在程序打印出最大的素数，假设合成数小于52，而对于大于52的范围，它不会打印出正确的输出。我附加了代码，请看你们能不能帮我|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2604143|#include<stdio.h>|code-block|syntax|javascript|2604144|void+main()+{+int+i，j，b=2，c；printf(“\n输入复合数字：")；scanf("%25d"，&c)；printf("Factors：")；|2604145|for(i=1;+i<=c/2;+i%2B%2B)
{
++++if(c%25i==0)
++++{
++++++++printf("%25d+",+i);
++++++++for(j=1;+j<=i;+j%2B%2B)+//since+a+numbr+cand+be+divisible+by+a+number+greated+than+its+half
++++++++{+++++++++++++++if(i%25j+>+0)
++++++++++++++++b+=+i;+//b+stores+the+largest+prime+factor
++++++++++++if(b%253==0)
++++++++++++++++b+=+3;
++++++++++++++++else+if(b%252==0)
++++++++++++++++b=2;
++++++++++++++else+if(b%255==0)
++++++++++++++++b=5;
++++++++++++++++}
++++}
}


printf("%25d\nLargest+prime+factor:+%25d\n",+c,+b);|2604146|}|2604147|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Hey thanks for the input friends, i worked out something and now the program prints the greatest prime factor of the composite number ,provided that the composite number is lesser than 52, while for higher ranges above this, it doesn't print the right output.
I'm Appending the code, please see if you guys could help me around

<pre><code>#include&lt;stdio.h&gt;
</code></pre>

void main()
{
 int i,j,b=2,c;
 printf("\nEnter a composite number: ");
 scanf("%d", &amp;c);
 printf("Factors: ");

<pre><code>for(i=1; i&lt;=c/2; i++)
{
 if(c%i==0)
 {
 printf("%d ", i);
 for(j=1; j&lt;=i; j++) //since a numbr cand be divisible by a number greated than its half
 { if(i%j &gt; 0)
 b = i; //b stores the largest prime factor
 if(b%3==0)
 b = 3;
 else if(b%2==0)
 b=2;
 else if(b%5==0)
 b=5;
 }
 }
}


printf("%d\nLargest prime factor: %d\n", c, b);
</code></pre>

}

blocks|key|182140|text|在Python中，类似这样的代码就可以了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|182141|import+math
import+itertools

#+largest+prime+factor+of+a+composite+number
def+primality(num):
++++for+i+in+range(2,int(math.sqrt(num))%2B1):
++++++++if+num%25i+==0:
+++++++++++++return+0
++++return+1

if+__name__+==+'__main__':
++++number+=+600851475143
++++for+i+in+itertools.count(2,1):
++++++++++++if+number%25i+==+0:
++++++++++++++++if+number%2510+in+[7,9,1,3]+and+primality(number/i):
++++++++++++++++++++print+number/i
++++++++++++++++++++break|code-block|syntax|javascript|182142|我所做的检查素数是整洁的平方根技术。|182143|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In Python, Something like this will do:

<pre><code>import math
import itertools

# largest prime factor of a composite number
def primality(num):
 for i in range(2,int(math.sqrt(num))+1):
 if num%i ==0:
 return 0
 return 1

if __name__ == '__main__':
 number = 600851475143
 for i in itertools.count(2,1):
 if number%i == 0:
 if number%10 in [7,9,1,3] and primality(number/i):
 print number/i
 break
</code></pre>

What I have done to check primality is the neat square root technique.

I was solving a puzzle, where im required to find the largest Prime Factor of a composite number entered by the user.
I thought of something and have tried it out, but it doesn't manage to detect the largest prime factor amongst the factors of the composite number.

I'm appending my code below, I'd be grateful if anyone could help me out here to get to detect the largest prime no. amongst the factors and print it. 

<pre><code>// Accept a composite number from user and print its largest prime factor.

#include&lt;stdio.h&gt;
void main()
{
 int i,j,b=2,c;
 printf("\nEnter a composite number: ");
 scanf("%d", &amp;c);
 printf("Factors: ");

 for(i=1; i&lt;=c/2; i++)
 {
 if(c%i==0)
 {
 printf("%d ", i);
 for(j=2; j&lt;=i/2; j++) //since a numbr cand be divisible by a number greated than its half
 { if(i%j &gt; 0) 
 b = i;
 else if(i==3)
 b = 3;
 }
 }
 }
 printf("%d\nLargest prime factor: %d\n", c, b);
}
</code></pre>

Printing the Largest Prime Factor of a Composite Number in C

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在解决一个难题，其中我需要找到用户输入的合成数的最大素数因子。我想到了一些东西，并进行了尝试，但它无法检测到合成数的因子中最大的素因数。我附上了下面的代码，如果有人能帮我找出最大的素数no，我将不胜感激。并将其打印出来。// Accept a composite number from user and print its largest prime factor.#include<stdio

问用C语言打印合成数的最大素因数
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回答 5

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