[英]Find the largest palindrome made from the product of two 3-digit numbers. (Java)
[英]Find the largest palindrome made from the product of two 3-digit numbers
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int value = 0;
for(int i = 100;i <=999;i++)
{
for(int j = i;j <=999;j++)
{
int value1 = i * j;
StringBuilder sb1 = new StringBuilder(""+value1);
String sb2 = ""+value1;
sb1.reverse();
if(sb2.equals(sb1.toString()) && value<value1) {
value = value1;
}
}
}
System.out.println(value);
}
}
這是我用Java編寫的代碼......除此之外還有什么有效的方法..我們可以更優化這些代碼嗎?
我們假設最大的這樣的回文將有六位而不是五位,因為143 * 777 = 111111是回文。
如其他地方所述,6位數的10-palindrome abccba是11的倍數。這是正確的,因為* 100001 + b * 010010 + c * 001100等於11 * a * 9091 + 11 * b * 910 + 11 * C * 100。 因此,在我們的內循環中,如果m不是11的倍數,我們可以將n減少11步。
我們正試圖找到百萬以下最大的回文,這是兩個3位數字的乘積。 為了找到一個大的結果,我們首先嘗試大的除數:
我們追蹤到目前為止變量q中發現的最大回文。 假設q = r·s,r <= s。 我們通常有m <r <= s。 我們要求m·n> q或n> = q / m。 由於發現較大的回文,n的范圍受到更多限制,原因有兩個:q變大,m變小。
附加程序的內部循環僅執行506次,相對於所使用的朴素程序的~810000次。
#include <stdlib.h>
#include <stdio.h>
int main(void) {
enum { A=100000, B=10000, C=1000, c=100, b=10, a=1, T=10 };
int m, n, p, q=111111, r=143, s=777;
int nDel, nLo, nHi, inner=0, n11=(999/11)*11;
for (m=999; m>99; --m) {
nHi = n11; nDel = 11;
if (m%11==0) {
nHi = 999; nDel = 1;
}
nLo = q/m-1;
if (nLo < m) nLo = m-1;
for (n=nHi; n>nLo; n -= nDel) {
++inner;
// Check if p = product is a palindrome
p = m * n;
if (p%T==p/A && (p/B)%T==(p/b)%T && (p/C)%T==(p/c)%T) {
q=p; r=m; s=n;
printf ("%d at %d * %d\n", q, r, s);
break; // We're done with this value of m
}
}
}
printf ("Final result: %d at %d * %d inner=%d\n", q, r, s, inner);
return 0;
}
注意,程序是在C中,但相同的技術將在Java中工作。
我會怎么做:
一些代碼:
int[] primes = new int[] {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,,127,131,137,139,149,151,157,163,167,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,
947,953,967,971,977,983,991,997};
for(int i = 999; i >= 100; i--) {
String palstr = String.valueOf(i) + (new StringBuilder().append(i).reverse());
int pal = Integer.parseInt(pal);
int[] factors = new int[20]; // cannot have more than 20 factors
int remainder = pal;
int facpos = 0;
primeloop:
for(int p = 0; p < primes.length; i++) {
while(remainder % p == 0) {
factors[facpos++] = p;
remainder /= p;
if(remainder < p) break primeloop;
}
}
// now to do the combinations here
}
我們可以將任務翻譯成數學語言。
首先,我們使用字符作為數字:
abc * xyz = n
abc is a 3-digit number, and we deconstruct it as 100*a+10*b+c
xyz is a 3-digit number, and we deconstruct it as 100*x+10*y+z
現在我們有兩個數學表達式,可以將a,b,c,x,y,z定義為{0..9}的€。
從{1..9}定義a和x作為元素更精確,而不是{0..9},因為097實際上不是一個3位數,是嗎?
好。
如果我們想要產生一個大數字,我們應該嘗試達到9 ......-數字,因為它應該是回文,它必須是9 ... 9的模式。 如果最后一個數字是9,那么從
(100*a + 10*b + c) * (100*x + 10*y + z)
如果z * c必須導致一個數字,以數字9結尾 - 所有其他計算都不會感染最后一個數字。
所以c和z必須來自(1,3,7,9),因為(1 * 9 = 9,9 * 1 = 9,3 * 3 = 9,7 * 7 = 49)。
現在一些代碼(Scala):
val n = (0 to 9)
val m = n.tail // 1 to 9
val niners = Seq (1, 3, 7, 9)
val highs = for (a <- m;
b <- n;
c <- niners;
x <- m;
y <- n;
z <- niners) yield ((100*a + 10*b + c) * (100*x + 10*y + z))
然后我會按大小對它們進行排序,從最大的那個開始,測試它們是否為回文。 所以我會省略測試小數字的回文,因為這可能不那么便宜。
出於美學原因,我不會采用(toString.reverse == toString)方法,而是遞歸除法和模解決方案,但在今天的機器上,它沒有太大的區別,是嗎?
// Make a list of digits from a number:
def digitize (z: Int, nums : List[Int] = Nil) : List[Int] =
if (z == 0) nums else digitize (z/10, z%10 :: nums)
/* for 342243, test 3...==...3 and then 4224.
Fails early for 123329 */
def palindromic (nums : List[Int]) : Boolean = nums match {
case Nil => true
case x :: Nil => true
case x :: y :: Nil => x == y
case x :: xs => x == xs.last && palindromic (xs.init) }
def palindrom (z: Int) = palindromic (digitize (z))
出於嚴格的性能考慮,我將針對toString / reverse / equals方法進行測試。 也許情況更糟。 它應該盡早失敗,但是不知道除法和模數是最快的操作,我用它們從Int創建一個List。 它適用於BigInt或Long,幾乎沒有重新聲明,並且適用於Java; 可以用Java實現,但在那里看起來不同。
好吧,把事情放在一起:
highs.filter (_ > 900000) .sortWith (_ > _) find (palindrom)
res45: Option[Int] = Some(906609)
那里有835個數字> 900000,並且它返回的速度相當快,但我想更強大的蠻力也不會慢得多。
也許有一種更聰明的方法來構建最高的palindrom,而不是搜索它。
一個問題是:我以前不知道,有一個解決方案> 900000。
一種非常不同的方法是產生大的回文,並解構它們的因素。
public class Pin
{
public static boolean isPalin(int num)
{
char[] val = (""+num).toCharArray();
for(int i=0;i<val.length;i++)
{
if(val[i] != val[val.length - i - 1])
{
return false;
}
}
return true;
}
public static void main(String[] args)
{
for(int i=999;i>100;i--)
for(int j=999;j>100;j--)
{
int mul = j*i;
if(isPalin(mul))
{
System.out.printf("%d * %d = %d",i,j,mul);
return;
}
}
}
}
package ex;
public class Main {
public static void main(String[] args) {
int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, flag = 0;
for (i = 999; i >= 100; i--) {
for (j = i; j >= 100; j--) {
k = i * j;
// System.out.println(k);
m = 0;
n = k;
while (n > 0) {
l = n % 10;
m = m * 10 + l;
n = n / 10;
}
if (m == k) {
System.out.println("pal " + k + " of " + i + " and" + j);
flag = 1;
break;
}
}
if (flag == 1) {
// System.out.println(k);
break;
}
}
}
}
一種稍微不同的方法,可以輕松計算最多兩個6位數字的乘積所產生的最大回文數。
第一部分是創建回文數的發生器。 所以沒有必要檢查一個數字是否是回文,第二部分是一個簡單的循環。
#include <memory>
#include <iostream>
#include <cmath>
using namespace std;
template <int N>
class PalindromeGenerator {
unique_ptr <int []> m_data;
bool m_hasnext;
public :
PalindromeGenerator():m_data(new int[N])
{
for(auto i=0;i<N;i++)
m_data[i]=9;
m_hasnext=true;
}
bool hasNext() const {return m_hasnext;}
long long int getnext()
{
long long int v=0;
long long int b=1;
for(int i=0;i<N;i++){
v+=m_data[i]*b;
b*=10;
}
for(int i=N-1;i>=0;i--){
v+=m_data[i]*b;
b*=10;
}
auto i=N-1;
while (i>=0)
{
if(m_data[i]>=1) {
m_data[i]--;
return v;
}
else
{
m_data[i]=9;
i--;
}
}
m_hasnext=false;
return v;
}
};
template<int N>
void findmaxPalindrome()
{
PalindromeGenerator<N> gen;
decltype(gen.getnext()) minv=static_cast<decltype(gen.getnext())> (pow(10,N-1));
decltype(gen.getnext()) maxv=static_cast<decltype(gen.getnext())> (pow(10,N)-1);
decltype(gen.getnext()) start=11*(maxv/11);
while(gen.hasNext())
{
auto v=gen.getnext();
for (decltype(gen.getnext()) i=start;i>minv;i-=11)
{
if (v%i==0)
{
auto r=v/i;
if (r>minv && r<maxv ){
cout<<"done:"<<v<<" "<<i<< "," <<r <<endl;
return ;
}
}
}
}
return ;
}
int main(int argc, char* argv[])
{
findmaxPalindrome<6>();
return 0;
}
i did this my way , but m not sure if this is the most efficient way of doing this .
package problems;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P_4 {
/**
* @param args
* @throws IOException
*/
static int[] arry = new int[6];
static int[] arry2 = new int[6];
public static boolean chk()
{
for(int a=0;a<arry.length;a++)
if(arry[a]!=arry2[a])
return false;
return true;
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int temp,z,i;
for(int x=999;x>100;x--)
for(int y=999;y>100;y--)
{
i=0;
z=x*y;
while(z>0)
{
temp=z%10;
z=z/10;
arry[i]=temp;
i++;
}
for(int k = arry.length;k>0;k--)
arry2[arry.length- k]=arry[k-1];
if(chk())
{
System.out.print("pelindrome = ");
for(int l=0;l<arry2.length;l++)
System.out.print(arry2[l]);
System.out.println(x);
System.out.println(y);
}
}
}
}
這是C中的代碼,有點長,但是完成了工作。:)
#include <stdio.h>
#include <stdlib.h>
/*
A palindromic number reads the same both ways. The largest palindrome made from the product of two
2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.*/
int palndr(int b)
{
int *x,*y,i=0,j=0,br=0;
int n;
n=b;
while(b!=0)
{
br++;
b/=10;
}
x=(int *)malloc(br*sizeof(int));
y=(int *)malloc(br*sizeof(int));
int br1=br;
while(n!=0)
{
x[i++]=y[--br]=n%10;
n/=10;
}
int ind = 1;
for(i=0;i<br1;i++)
if(x[i]!=y[i])
ind=0;
free(x);
free(y);
return ind;
}
int main()
{
int i,cek,cekmax=1;
int j;
for(i=100;i<=999;i++)
{
for(j=i;j<=999;j++)
{
cek=i*j;
if(palndr(cek))
{
if(pp>cekmax)
cekmax=cek;
}
}
}
printf("The largest palindrome is: %d\n\a",cekmax);
}
你可以用Python實際做到這一點,它很容易看看:
actualProduct = 0
highestPalindrome = 0
# Setting the numbers. In case it's two digit 10 and 99, in case is three digit 100 and 999, etc.
num1 = 100
num2 = 999
def isPalindrome(number):
number = str(number)
reversed = number[::-1]
if number==reversed:
return True
else:
return False
a = 0
b = 0
for i in range(num1,num2+1):
for j in range(num1,num2+1):
actualProduct = i * j
if (isPalindrome(actualProduct) and (highestPalindrome < actualProduct)):
highestPalindrome = actualProduct
a = i
b = j
print "Largest palindrome made from the product of two %d-digit numbers is [ %d ] made of %d * %d" % (len(str(num1)), highestPalindrome, a, b)
由於我們沒有同時循環使用迭代器(num1和num2),因此我們找到的第一個回文數將是最大的。 我們不需要測試我們發現的回文是否是最大的。 這大大減少了計算所需的時間。
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int limit = 99;
int max = 999;
int num1 = max, num2, prod;
while(num1 > limit)
{
num2 = num1;
while(num2 > limit)
{
total = num1 * num2;
StringBuilder sb1 = new StringBuilder(""+prod);
String sb2 = ""+prod;
sb1.reverse();
if( sb2.equals(sb1.toString()) ) { //optimized here
//print and exit
}
num2--;
}
num1--;
}
}//end of main
}//end of class PalindromeThreeDigits
我嘗試了Tobin joy和vickyhacks的解決方案,他們兩個都產生了結果580085這是錯的,這是我的解決方案,雖然非常笨拙:
import java.util.*;
class ProjEu4
{
public static void main(String [] args) throws Exception
{
int n=997;
ArrayList<Integer> al=new ArrayList<Integer>();
outerloop:
while(n>100){
int k=reverse(n);
int fin=n*1000+k;
al=findfactors(fin);
if(al.size()>=2)
{
for(int i=0;i<al.size();i++)
{
if(al.contains(fin/al.get(i))){
System.out.println(fin+" factors are:"+al.get(i)+","+fin/al.get(i));
break outerloop;}
}
}
n--;
}
}
private static ArrayList<Integer> findfactors(int fin)
{
ArrayList<Integer> al=new ArrayList<Integer>();
for(int i=100;i<=999;i++)
{
if(fin%i==0)
al.add(i);
}
return al;
}
private static int reverse(int number)
{
int reverse = 0;
while(number != 0){
reverse = (reverse*10)+(number%10);
number = number/10;
}
return reverse;
}
}
最有可能的是復制其他解決方案之一但由於pythonified代碼它看起來很簡單,即使它有點蠻力。
def largest_palindrome():
largest_palindrome = 0;
for i in reversed(range(1,1000,1)):
for j in reversed(range(1, i+1, 1)):
num = i*j
if check_palindrome(str(num)) and num > largest_palindrome :
largest_palindrome = num
print "largest palindrome ", largest_palindrome
def check_palindrome(term):
rev_term = term[::-1]
return rev_term == term
怎么樣:在python中
>>> for i in range((999*999),(100*100), -1):
... if str(i) == str(i)[::-1]:
... print i
... break
...
997799
>>>
我相信有一種更簡單的方法:檢查從兩位三位數的最大乘積下降的回文,選擇具有兩個三位數因子的第一個回文。
這是Ruby代碼:
require './palindrome_range'
require './prime'
def get_3_digit_factors(n)
prime_factors = Prime.factors(n)
rf = [prime_factors.pop]
rf << prime_factors.shift while rf.inject(:*) < 100 || prime_factors.inject(:*) > 999
lf = prime_factors.inject(:*)
rf = rf.inject(:*)
lf < 100 || lf > 999 || rf < 100 || rf > 999 ? [] : [lf, rf]
end
def has_3_digit_factors(n)
return !get_3_digit_factors(n).empty?
end
pr = PalindromeRange.new(0, 999 * 999)
n = pr.downto.find {|n| has_3_digit_factors(n)}
puts "Found #{n} - Factors #{get_3_digit_factors(n).inspect}, #{Prime.factors(n).inspect}"
prime.rb:
class Prime
class<<self
# Collect all prime factors
# -- Primes greater than 3 follow the form of (6n +/- 1)
# Being of the form 6n +/- 1 does not mean it is prime, but all primes have that form
# See http://primes.utm.edu/notes/faq/six.html
# -- The algorithm works because, while it will attempt non-prime values (e.g., (6 *4) + 1 == 25),
# they will fail since the earlier repeated division (e.g., by 5) means the non-prime will fail.
# Put another way, after repeatedly dividing by a known prime, the remainder is itself a prime
# factor or a multiple of a prime factor not yet tried (e.g., greater than 5).
def factors(n)
square_root = Math.sqrt(n).ceil
factors = []
while n % 2 == 0
factors << 2
n /= 2
end
while n % 3 == 0
factors << 3
n /= 3
end
i = 6
while i < square_root
[(i - 1), (i + 1)].each do |f|
while n % f == 0
factors << f
n /= f
end
end
i += 6
end
factors << n unless n == 1
factors
end
end
end
palindrome_range.rb:
class PalindromeRange
FIXNUM_MAX = (2**(0.size * 8 -2) -1)
def initialize(min = 0, max = FIXNUM_MAX)
@min = min
@max = max
end
def downto
return enum_for(:downto) unless block_given?
n = @max
while n >= @min
yield n if is_palindrome(n)
n -= 1
end
nil
end
def each
return upto
end
def upto
return enum_for(:downto) unless block_given?
n = @min
while n <= @max
yield n if is_palindrome(n)
n += 1
end
nil
end
private
def is_palindrome(n)
s = n.to_s
i = 0
j = s.length - 1
while i <= j
break if s[i] != s[j]
i += 1
j -= 1
end
i > j
end
end
public class ProjectEuler4 {
public static void main(String[] args) {
int x = 999; // largest 3-digit number
int largestProduct = 0;
for(int y=x; y>99; y--){
int product = x*y;
if(isPalindormic(x*y)){
if(product>largestProduct){
largestProduct = product;
System.out.println("3-digit numbers product palindormic number : " + x + " * " + y + " : " + product);
}
}
if(y==100 || product < largestProduct){y=x;x--;}
}
}
public static boolean isPalindormic(int n){
int palindormic = n;
int reverse = 0;
while(n>9){
reverse = (reverse*10) + n%10;
n=n/10;
}
reverse = (reverse*10) + n;
return (reverse == palindormic);
}
}
您可以使用11是回文的倍數來減少搜索空間的事實。 我們可以得到這個,因為我們可以假設回文將是6位數並且> = 111111。
例如(來自projecteuler;))
P= xyzzyx = 100000x + 10000y + 1000z + 100z + 10y +x
P=100001x+10010y+1100z
P=11(9091x+910y+100z)
檢查i mod 11!= 0,然后j循環可以減去11(從990開始),因為兩者中的至少一個必須可被11整除。
您可以嘗試以下打印
999 * 979 * 989 = 967262769
largest palindrome= 967262769 took 0.015
public static void main(String... args) throws IOException, ParseException {
long start = System.nanoTime();
int largestPalindrome = 0;
for (int i = 999; i > 100; i--) {
LOOP:
for (int j = i; j > 100; j--) {
for (int k = j; k > 100; k++) {
int n = i * j * k;
if (n < largestPalindrome) continue LOOP;
if (isPalindrome(n)) {
System.out.println(i + " * " + j + " * " + k + " = " + n);
largestPalindrome = n;
}
}
}
}
long time = System.nanoTime() - start;
System.out.printf("largest palindrome= %d took %.3f seconds%n", largestPalindrome, time / 1e9);
}
private static boolean isPalindrome(int n) {
if (n >= 100 * 1000 * 1000) {
// 9 digits
return n % 10 == n / (100 * 1000 * 1000)
&& (n / 10 % 10) == (n / (10 * 1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (1000 * 1000) % 10)
&& (n / 1000 % 10) == (n / (100 * 1000) % 10);
} else if (n >= 10 * 1000 * 1000) {
// 8 digits
return n % 10 == n / (10 * 1000 * 1000)
&& (n / 10 % 10) == (n / (1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (100 * 1000) % 10)
&& (n / 1000 % 10) == (n / (10 * 1000) % 10);
} else if (n >= 1000 * 1000) {
// 7 digits
return n % 10 == n / (1000 * 1000)
&& (n / 10 % 10) == (n / (100 * 1000) % 10)
&& (n / 100 % 10) == (n / (10 * 1000) % 10);
} else throw new AssertionError();
}
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