c ++阶乘数中的零数 [英] c++ Number of zeros in a factorial number

问题描述

可能重复：

它给出一个整数p。我必须找到一个数字n，其中n阶乘在末尾有p个数字为零。这里是我想的解决方案，但我不知道如果它是这个问题的解决方案。

  int p; 
 int count5 = 0; 
 int i; 
 int copy_i; 
 
 printf（Enter p：）; 
 scanf（％d，& p）; 
 
 for（i = 1;; i ++）
 {
 copy_i = i; 
 
 while（copy_i / 5）
 {
 if（copy_i％5 == 0）
 {
 count5 ++; 
 copy_i = copy_i / 5; 
} 
 else 
 {
 break; 
} 
} 
 
 if（count5 == p）
 {
 printf（最小数字n是：％d。 ; 
 break; 
} 
 else if（count5> p）
 {
 printf（No match for n！with％d zero。 
 break; 
} 
 
} 
  

解决方案<

这听起来像一个Project Euler问题，所以我不会给出明确的解决方案。这里有两个提示：

• 你不必实际计算阶乘以找出结尾有多少个零。 li>
  
• 如果数字可被2 ^ N和5 ^ N除尽，则该数字在结尾处至少有N个零。

Possible Duplicate:
Counting trailing zeros of numbers resulted from factorial

It is given an integer "p". I have to find a number "n" for which "n factorial" has "p" numbers of zero at the end. Here is the solution i thought, but i am not sure if it is a solution for this problem. Do i have to make a function to calculate the factorial and another function to get the zeros?

int p;
int count5=0;
int i;
int copy_i;

printf("Enter p: ");
scanf("%d",&p);

for(i=1; ;i++)
{
    copy_i=i;

    while(copy_i/5)
    {
        if(copy_i%5==0)
        {
            count5++;
            copy_i=copy_i/5;
        }
        else
        {
            break;
        }
    }

    if(count5==p)
    {
        printf("The minimum number n is: %d.",i);
        break;
    }
    else if(count5>p) 
    {
        printf("No match for n! with %d zero.",p);
        break;
    }

}

解决方案

This sounds like a Project Euler problem, so I won't give an explicit solution. Here's two hints though:

• You do not have to actually calculate the factorial to find out how many zeroes it has at the end.
• If a number is divisible by 2^N and by 5^N, the number has at least N zeroes at the end.

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