求階乘的第一個(gè)非零數(shù)字

Problem statement:

問題陳述：

Find the number of trailing zeros in n! (Where, n is the given input).

在n中找到尾隨零的數(shù)目！ (其中， n是給定的輸入)。

Solution:

解：

Computing a factorial is of course expansive. Though using dynamic programming the computing expanse can be managed, for the large value of n, the factorial value is going exceed normal data size. Needless to say, computing the whole factorial is not the way to find the number of trailing zeros. There must be some advanced algorithm to find the no of trailing zeros.

計(jì)算階乘當(dāng)然是可擴(kuò)展的。 盡管使用動(dòng)態(tài)編程可以管理計(jì)算范圍，但是對于較大的n值，階乘值將超過正常數(shù)據(jù)大小。 不用說，計(jì)算整個(gè)階乘不是找到尾隨零的數(shù)量的方法 。 必須有一些高級(jí)算法來找到尾隨零 。

Firstly, we need to understand what causes trailing zeroes. A pair of 2 & 5 is the reason behind a trailing zero. Thus a pair of 2 & 5 in the factorial expression leads to a trailing zero. Thus we simply need to check how many pairs (different) of 2 & 5 are there.

首先，我們需要了解導(dǎo)致尾隨零的原因。 一對2和5是尾隨零后面的原因。 因此，階乘表達(dá)式中的2和5對導(dǎo)致尾隨零。 因此，我們只需要檢查2和5有多少對(不同)。

Let's say 5!
5!= 5*4*3*2*1 (thus only one pair)
5!=120 ( only one trailing zero)

假設(shè)5！
5！= 5 * 4 * 3 * 2 * 1 (因此只有一對)
5！= 120 (僅一個(gè)尾隨零)

Intuition says that we don’t even need to find the number of pairs as the occurrence of 2 as a factor is obvious if a 5 is present as a factor. Thus we only need to check how many 5 is there as a factor in the factorial.

直覺說，我們甚至不需要找到對的數(shù)量，因?yàn)槿绻?作為一個(gè)因素，則2作為一個(gè)因素的出現(xiàn)就很明顯。 因此，我們僅需要檢查階乘中有5個(gè)因素。

Algorithm:

算法：

Set count to 0For(i=5;n/i>0;i=i*5)count=count+ n/i;Return count .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

C ++代碼查找數(shù)字階乘中的尾隨零 (C++ code to find trailing zeros in factorial of a number)

#include<bits/stdc++.h> using namespace std;int trailingZeros(int n){int count=0;if(n<0)return -1;for(int i=5;n/i>0;i*=5){count+=n/i;}return count; }int main(){int n;cout<<"enter input,n"<<endl;cin>>n;if(trailingZeros(n))cout<<"no of trailing zero in "<<n<<"! is "<<trailingZeros(n)<<endl;elsecout<<"input is negative, can't proceed"<<endl;return 0; }

Output

輸出量

First run: enter input,n 15 no of trailing zero in 15! is 3Second run: enter input,n 100 no of trailing zero in 100! is 24

翻譯自: https://www.includehelp.com/algorithms/find-trailing-zeros-in-factorial-of-a-number.aspx

求階乘的第一個(gè)非零數(shù)字

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