[编程题]大整数的因子
- 热度指数：11920 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
- 算法知识视频讲解
已知正整数k满足2<=k<=9，现给出长度最大为30位的十进制非负整数c，求所有能整除c的k.
输入描述:
若干个非负整数c，c的位数<=30 每行一个c
输出描述:
每一个c的结果占一行 1) 若存在满足 c%k == 0 的k，输出所有这样的k，中间用空格隔开，最后一个k后面没有空格。 2) 若没有这样的k则输出"none" 注意整数溢出问题 不要对-1进行计算
示例1
输入
30 72 13 -1
输出
2 3 5 6 2 3 4 6 8 9 none
import java.util.Scanner;
import java.math.BigInteger;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
String str = in.next();
if (str.equals("-1")) {
break;
}
BigInteger c = new BigInteger(str);
boolean find = false;
for (int k = 2; k <= 9; k++) {
if (c.mod(new BigInteger(String.valueOf(k)))
.equals(new BigInteger("0"))) {
find = true;
System.out.print(k);
for (int k1 = k + 1; k1 <= 9; k1++) {
if (c.mod(new BigInteger(String.valueOf(k1)))
.equals(new BigInteger("0"))) {
System.out.print(" " + k1);
}
}
System.out.println();
break;
}
}
if (!find) {
System.out.println("none");
}
}
}
}
编辑于 2024-01-30 20:54:12
回复(0)
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
while(in.hasNext())
{
BigInteger b=in.nextBigInteger();
int len=0;
for(int j=2;j<=9;j++)
{
BigInteger t=BigInteger.valueOf(j);
if(b.remainder(t).equals(BigInteger.ZERO))
{
len++;
System.out.print(j+" ");
}
}
if(len==0)
{
System.out.println("none");
}else
{
System.out.println();
}
}
}
}
发表于 2020-04-16 20:54:11
回复(0)
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext()){
BigInteger n = in.nextBigInteger(); //从键盘获取一个大数值
int sum = 0;
int flag = 0;
int index = 0; //定义index表示是否存在被除尽的数
for(int i = 2; i <= 9; i++){
BigInteger a = new BigInteger(String.valueOf(i));
BigInteger num = n.mod(a);
BigInteger b = new BigInteger("0");
if(num.compareTo(b) == 0){
index = 1;
sum++; //定义sum记录一共有多少个值能被除尽
}
}
for(int i = 2; i <= 9; i++){
BigInteger a = new BigInteger(String.valueOf(i));
BigInteger num = n.mod(a);
BigInteger b = new BigInteger("0");
if(num.compareTo(b) == 0){
flag++;
System.out.print(flag != sum ? i + " " : i);
}
}
if(index == 0){
System.out.println("none");
}
}
in.close();
}
}
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext()){
BigInteger n = in.nextBigInteger(); //从键盘获取一个大数值
int sum = 0;
int flag = 0;
int index = 0; //定义index表示是否存在被除尽的数
for(int i = 2; i <= 9; i++){
BigInteger a = new BigInteger(String.valueOf(i));
BigInteger num = n.mod(a);
BigInteger b = new BigInteger("0");
if(num.compareTo(b) == 0){
index = 1;
sum++; //定义sum记录一共有多少个值能被除尽
}
}
for(int i = 2; i <= 9; i++){
BigInteger a = new BigInteger(String.valueOf(i));
BigInteger num = n.mod(a);
BigInteger b = new BigInteger("0");
if(num.compareTo(b) == 0){
flag++;
System.out.print(flag != sum ? i + " " : i);
}
}
if(index == 0){
System.out.println("none");
}
}
in.close();
}
}
发表于 2020-03-02 22:28:52
回复(0)
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
while (input.hasNext()) {
String c = input.nextLine();
boolean flag = true;
if ("-1".equals(c)) {
break;
} else {
BigInteger big = new BigInteger(c);
for (int i = 2; i <= 9; i++) {
if (big.mod(BigInteger.valueOf(i)).intValue() == 0) {
if (!flag) {
System.out.print(" ");
}
System.out.printf("%d", i);
flag = false;
}
}
}
if (flag) {
System.out.println("none");
} else {
System.out.println();
}
}
}
}
发表于 2020-02-28 09:39:59
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
while (reader.hasNext()) {
String input = reader.next();
ArrayList<Integer> res = new ArrayList<>();
for (int divisor = 2; divisor <= 9; ++divisor) {
int mod = 0;
for (int j = 0; j < input.length(); ++j) {
int digit = input.charAt(j) - '0';
mod = (mod * 10 + digit) % divisor;
}
if (mod == 0)
res.add(divisor);
}
StringBuilder sb = new StringBuilder();
if (res.size() > 0) {
for (int i: res) {
sb.append(i+" ");
}
System.out.println(sb.substring(0, sb.length()-1).toString());
} else {
System.out.println("none");
}
}
}
}
发表于 2018-04-07 22:27:03
回复(0)
直接用除法，思路就是上一位取余数*10加上当前位除以k，
一直到所有位判断完毕，这个其实就是手算除法的模拟过程
import java.util.Scanner;
/*
*
*直接用除法，思路就是上一位取余数*10加上当前位除以k，
*一直到所有位判断完毕，这个其实就是手算除法的模拟过程
*
* */
public class BigNumer {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
String num = in.nextLine();
if (num.equals("-1")) {
break;
}
process(num);
}
in.close();
}
static void process(String num) {
StringBuffer buffer = new StringBuffer();
for (int k = 2; k < 10; k++) {
if (isContaineFactor(num, k)) {
buffer.append(k + " ");
}
}
if (buffer.length() > 0) {
buffer.delete(buffer.length() - 1, buffer.length());
}
else {
buffer.append("none");
}
System.out.println(buffer.toString());
}
static boolean isContaineFactor(String num, int k) {
int num1 = 0;
int num2 = 0;
int len = num.length();
for (int i = 0; i < len; i++) {
num1 = (num.charAt(i) - '0') + num2 * 10;
num2 = num1 % k;
}
return num2 == 0 ? true : false;
}
}
发表于 2016-07-21 12:05:46
回复(0)
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