功能:输入一个正整数，按照从小到大的顺序输出它的所有质因子（重复的也要列举）（如180的质因子为2 2 3 3 5 ）
数据范围： 
[编程题]质数因子
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
long num = sc.nextLong();
long num2 = (long)Math.sqrt(num);
for(int i = 2; i <= num2; i++){
while(num % i == 0){
num /= i;
System.out.print(i + " ");
}
}
System.out.println(num == 1 ? "" : num + " ");
}
sc.close();
}
}
发表于 2021-08-16 18:53:16
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import java.util.*;
import java.io.*;
public class Main {
/*
既是质数又是因子
*/
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = "";
while ((str = br.readLine()) != null) {
long n = Long.parseLong(str);
String res = getStr(n);
System.out.println(res);
}
}
// 获取输出格式
public static String getStr(long num) {
String ret = "";
for(int i = 2; i*i <= num; i++) { // 求质数
while (num % i == 0) {// 求因子
ret += i + " ";
num /= i;
}
}
if (num != 1) ret += num + " ";
return ret;
}
}
import java.io.*;
public class Main {
/*
既是质数又是因子
*/
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = "";
while ((str = br.readLine()) != null) {
long n = Long.parseLong(str);
String res = getStr(n);
System.out.println(res);
}
}
// 获取输出格式
public static String getStr(long num) {
String ret = "";
for(int i = 2; i*i <= num; i++) { // 求质数
while (num % i == 0) {// 求因子
ret += i + " ";
num /= i;
}
}
if (num != 1) ret += num + " ";
return ret;
}
}
发表于 2021-07-16 17:29:51
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public static void main(String[] args){
Scanner sc = new Scanner(System.in);
long s = sc.nextLong();
int temp = 2;
StringJoiner str= new StringJoiner(" ");
while(temp<=s){
if(s%temp==0){
if(temp==s){
str.add(temp+"");
break;
}else{
str.add(temp+"");
s=s/temp;
}
} else {
temp++;
}
}
} 这个 是啥问题
发表于 2020-08-18 21:30:33
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public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
long num = scanner.nextLong();
if(num <= 3){
System.out.println(num);
continue;
}
for (int i = 2; i <= num; i++) {
while (num % i == 0) {
System.out.println(i + " ");
num /= i;
}
}
}
}
发表于 2020-08-08 13:50:08
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不考虑质因子的，全都是耍流氓
贴个C的吧
#include <stdio.h>
#define ARR_LEN 200
#define TABLE_LEN 10000
static char ans[ARR_LEN];
static long numTable[TABLE_LEN] = {2,};
static int numTableLen = 1;
char* getResult(long ulDataInput)
{
int lenAns = 0;
memset(ans,0,sizeof(char) * ARR_LEN);
while(1 != ulDataInput)
{
while(0 == ulDataInput%numTable[numTableLen - 1])
{
int end = lenAns;
long num = numTable[numTableLen - 1];
ans[end] = ' ';
end++;
while(num)
{
ans[end] = num%10 + '0';
num = num/10;
end++;
}
for(int i = 0; i < (end - lenAns)/2; i++)
{
char tmp = ans[lenAns + i];
ans[lenAns + i] = ans[end - 1 - i];
ans[end - 1 - i] = tmp;
}
lenAns = end;
ulDataInput /= numTable[numTableLen - 1];
}
if(ulDataInput > numTable[numTableLen - 1])
{
for(long i = numTable[numTableLen - 1] + 1; i <= ulDataInput; i++)
{
int flag = 1;
for(int j = 0; j < numTableLen; j++)
{
if(0 == i%numTable[j])
{
flag = 0;
break;
}
}
if(flag)
{
numTable[numTableLen] = i;
numTableLen++;
break;
}
}
}
}
ans[lenAns] = '\0';
return ans;
}
int main()
{
long n = 0;
scanf("%d",&n);
printf("%s\n",getResult(n));
return 0;
}
编辑于 2020-04-14 10:29:15
回复(0)
a = int(input())
Outlist = []
for i in range(2, a // 2 + 1):
#for i in range(2, int(math.sqrt(a)) + 1):
while a % i == 0:
a = a / i
Outlist.append(i)
Outlist = []
for i in range(2, a // 2 + 1):
#for i in range(2, int(math.sqrt(a)) + 1):
while a % i == 0:
a = a / i
Outlist.append(i)
print(" ".join(map(str, Outlist)) + " " if Outlist else str(a) + " ")
此题的关键是找出所有质数因子，而且每个质数因子可能有多个
发表于 2019-08-31 10:31:30
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//第一次用递归解决了问题，纪念一哈
#include<stdio.h>
int z(long);//判断质数 int z(long a)
{
for(int i=2;i<a;i++)
{
if(a%i==0)
return 0;
}
return 1;
}
void y(long);//递归找因子
void y(long a)
{
int k=2;
while(a%k!=0&&k<a){
k++;
}
if(z(k)==1) printf("%d ",k);
if(k<a) y(a/k);
}
int main()
{
long x;
while(scanf("%d",&x)!=EOF){
y(x);
}
return 0;
}
编辑于 2018-08-18 17:00:12
回复(0)
高票答案没有忽略了题目要求的字符串转化...
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int hex2dec(string s);
string getResult(long ulDataInput);
string myitoa(long a);
int main(){
long str;
string a;
while (cin >> str)
cout << getResult(str);
return 0;
}
string getResult(long ulDataInput)
{
string str = "";
char a[100];
for (int i = 2; i <= ulDataInput; )
{
if ((ulDataInput%i) == 0)
{
ulDataInput /= i;
str += myitoa(i);
str += " ";
}
else
i++;
}
return str;
}
string myitoa(long a)
{
string str="";
vector<int> nums_r;
vector<int> nums;
while (a)
{
nums_r.push_back(a % 10);
a /= 10;
}
for (auto i = nums_r.end()-1;i>nums_r.begin();i--)
nums.push_back(*i);
nums.push_back(nums_r[0]);
for (int i : nums)
{
switch (i)
{
case 0: str += "0"; break;
case 1: str += "1"; break;
case 2: str += "2"; break;
case 3: str += "3"; break;
case 4: str += "4"; break;
case 5: str += "5"; break;
case 6: str += "6"; break;
case 7: str += "7"; break;
case 8: str += "8"; break;
case 9: str += "9"; break;
default:
break;
}
}
return str;
}
#include <string>
#include <vector>
using namespace std;
int hex2dec(string s);
string getResult(long ulDataInput);
string myitoa(long a);
int main(){
long str;
string a;
while (cin >> str)
cout << getResult(str);
return 0;
}
string getResult(long ulDataInput)
{
string str = "";
char a[100];
for (int i = 2; i <= ulDataInput; )
{
if ((ulDataInput%i) == 0)
{
ulDataInput /= i;
str += myitoa(i);
str += " ";
}
else
i++;
}
return str;
}
string myitoa(long a)
{
string str="";
vector<int> nums_r;
vector<int> nums;
while (a)
{
nums_r.push_back(a % 10);
a /= 10;
}
for (auto i = nums_r.end()-1;i>nums_r.begin();i--)
nums.push_back(*i);
nums.push_back(nums_r[0]);
for (int i : nums)
{
switch (i)
{
case 0: str += "0"; break;
case 1: str += "1"; break;
case 2: str += "2"; break;
case 3: str += "3"; break;
case 4: str += "4"; break;
case 5: str += "5"; break;
case 6: str += "6"; break;
case 7: str += "7"; break;
case 8: str += "8"; break;
case 9: str += "9"; break;
default:
break;
}
}
return str;
}
发表于 2018-07-07 21:30:52
回复(0)
import java.util.Scanner;
//渣渣的我用渣渣的方法，比不上其他大神
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
while(input.hasNext()) {
System.out.print(getResult(input.nextLong()));
}
}
public static String getResult(long ulDataInput) {
long divNum = 2;
String result = "";
while(divNum <= ulDataInput) {
if(ulDataInput > 1 && (ulDataInput%divNum == 0)) {
ulDataInput /=divNum;
result = result+divNum+" ";
}else {
divNum++;
}
if(ulDataInput == 1)break;
}
if(result.length()==0)
result = result+ulDataInput+" ";
return result;
}
//利用Java判断一个数是否是素数的算法
static boolean f(long a){
boolean ean = true;
for(int i=2;i< Math.sqrt(a);i++){ //Math.sqrt 是调用Math类中的sqrt方法，求一个数的平方根
if(a%i==0){
ean = false;
break;
}
}
return ean;
}
}
发表于 2018-05-23 22:22:46
回复(0)
#include<iostream>
#include<math.h>
#include<math.h>
using namespace std;
void getResult(long ulDataInput) {
long i = 1;
while (ulDataInput != 1 && i != ceil(sqrt(ulDataInput)))
{
for (i = 2; i <= sqrt(ulDataInput); ++i)
{
if (ulDataInput % i == 0)
{
ulDataInput /= i;
cout << i << " ";
break;
}
}
}
cout << ulDataInput << " " << endl;
}
long i = 1;
while (ulDataInput != 1 && i != ceil(sqrt(ulDataInput)))
{
for (i = 2; i <= sqrt(ulDataInput); ++i)
{
if (ulDataInput % i == 0)
{
ulDataInput /= i;
cout << i << " ";
break;
}
}
}
cout << ulDataInput << " " << endl;
}
int main() {
long ulDataInput;
long ulDataInput;
while (cin >> ulDataInput)
getResult(ulDataInput);
getResult(ulDataInput);
return 0;
}
}
发表于 2018-02-11 21:10:48
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//package com.gulamjan.t021.华为质数因子;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
long n = in.nextLong();
System.out.println(getPrimeString(n));
}
in.close();
}
//判断 n 是否是个质数
public static boolean isPrimeNumber(long n) {
boolean flag = true;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
flag = false;
break;
}
}
return flag;
}
//获取符合要求的因子字符串
public static String getPrimeString(long n) {
long temp = n;
String string = "";
for (int i = 2; i <= temp; i++) {
if (isPrimeNumber(i)) {
while(n % i == 0){
string += i+" ";
n = n / i;
}
}
}
return string;
}
}
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
long n = in.nextLong();
System.out.println(getPrimeString(n));
}
in.close();
}
//判断 n 是否是个质数
public static boolean isPrimeNumber(long n) {
boolean flag = true;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
flag = false;
break;
}
}
return flag;
}
//获取符合要求的因子字符串
public static String getPrimeString(long n) {
long temp = n;
String string = "";
for (int i = 2; i <= temp; i++) {
if (isPrimeNumber(i)) {
while(n % i == 0){
string += i+" ";
n = n / i;
}
}
}
return string;
}
}
发表于 2017-10-05 16:42:04
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#include <stdio.h>
#include <string.h>
#define N 100
int main()
{ long i; int j; long n; char over; long devisor[N]; scanf("%ld", &n); j = 0; while(1){ over = 1; for(i=2 ; i<=(n+1)/2 ; i++){ if(n%i == 0){ devisor[j++] = i; n /= i; over = 0; break; } } if(over) break; } for(i=0 ; i<j ; i++){ printf("%ld ", devisor[i]); } printf("%ld \n", n); }
编辑于 2017-09-17 10:54:27
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException{
BufferedReader bufferedReader = new
BufferedReader(new InputStreamReader(System.in));
Integer num = Integer.valueOf(bufferedReader.readLine());
//System.out.println("num = " + num);
for (int i = 2; i <= num; i++){
if (num % i == 0){
System.out.print(i + " ");
num = num / i; //重新赋值
i = 1; //重新遍历
}
}
bufferedReader.close();
}
}
编辑于 2017-09-15 13:58:24
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//java
//循环与判断
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNextLong()) {
Long a = sc.nextLong();
exchange(a);
}
}
public static void exchange(Long in) {
Long number = in;
while (number != 1) {
for (int i = 2; i <= number; i++) {
if (number % i == 0) {
number /= i;
System.out.print(i + " ");
break;
}
}
}
}
}
发表于 2017-09-08 16:41:56
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问题信息
通过挑战的用户
查看代码-
2023-03-07 20:23:25 -
2023-02-02 23:55:15
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2023-01-24 12:08:15 -
2023-01-07 14:37:29 -
2022-12-02 16:24:57
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