我的答案:function zeros (n) { var num=0; for(var i=1;i<=n;i++){ var j=i; while(j%5==0){ num=num+1; j=j/5; } } return num; }优秀答案:1 function zeros (n) { 2 var zs = 0; 3 while(n>0){ 4 n=Math.floor(n/5); 5 zs+=n 6 } 7 return zs; 8 }分类: c...
// Java program to illustrate the// Java.lang.Integer.numberOfTrailingZeros() methodimportjava.lang.*;publicclassTrailingZeros{publicstaticvoidmain(String[] args){inta =155; System.out.println("Integral Number = "+ a);// Returns the number of zero bits following the lowest-order//rightmost ...
zeros(12) = 2 # 123 .. 12 = 479001600 that has 2 trailing zeros 4790016(00) Be careful 1000! has length of 2568 digital numbers. My solution 只有当有2*5出现的时候,末尾才有可能出现0,而2的数量远大于5,所以我们只需要计算在N!中,有多少个5. function zeros (n) { var num = 0; while...
下麵的例子展示了 java.lang.Long.numberOfTrailingZeros() 方法的用法。 package com.tutorialspoint; import java.lang.*; public class LongDemo { public static void main(String[] args) { long l = 210; System.out.println("Number = " + l); /* returns the string representation of the unsigned...
java.lang.Integer.numberOfTrailingZeros() 方法返回最低位后面的零位数("最右端") ") 指定 int 值的二进制补码二进制表示形式中的一位。如果指定值的二进制补码表示形式中没有一位,即等于零,则该方法返回 32。 语法 public static int numberOfTrailingZeros(int i) 1 参数 i 指定要计算尾随零个数的值...
Integer.numberOfTrailingZeros() Method in Java with Example Java.lang.Integer.numberOfTrailingZeros() 是返回二进制中最低位(即最右边或最低有效“1”位)一位之后的零(0)位总数的方法补码指定整数值的二进制表示,或者我们可以说它是将 int 值转换为二进制然后考虑最低一位并返回 no 的函数。紧随其后的零...
NumberOfTrailingZeros ParseInt ParseUnsignedInt RemainderUnsigned Reverse ReverseBytes RotateLeft RotateRight Signum Sum ToBinaryString ToHexString ToOctalString ToString ToUnsignedLong ToUnsignedString ValueOf 运算符 显式接口实现 InternalError InterruptedException ...
in the position of the lowest-order ("rightmost") one-bit in the specified long value.*/System.out.println("Lowest one bit = "+Long.lowestOneBit(i));/*returns the number of zero bits preceding the highest-order ("leftmost")one-bit */System.out.print("Number of trailing zeros = ")...
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...
Integer: numberOfTrailingZeros(int i) importjava.io.IOException;publicclassMainClass {publicstaticvoidmain(String args[])throwsIOException {intn = 170;// 10101010System.out.println("Value in binary: 10101010"); System.out.println("Number of one bits: "+ Integer.bitCount(n)); System.out.print...