题目4：找出由两个三位数乘积构成的回文

永恒的蓝色梦想 · 发表于 2020-8-10 22:34:54

KingFish-v- 发表于 2020-8-10 21:44
答案是993*913=906609，用时0.01s左右。

找4位数x4位数的最大回文用时也在0.01s内，5位数x5位数用时0.58 ...

不错，我也学习一下

yhhpf · 发表于 2020-8-21 16:37:59

a = 1000

b = 1000

l1 = []

while a > 99:

a -= 1

b = 1000

while b >99:

      b -= 1

      s = a*b

      c = list(str(s))

      i = len(c)

      x = 1

      while x <= int(i/2) :

         if c[x-1] != c[0-x]:

            break

         if x == int(i/2):

            l1.append(s)

            break

         x += 1

print(max(l1))

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基本解法，忙完回头看~~~

a315734685 · 发表于 2020-9-1 17:19:43

# 一个回文数指的是从左向右和从右向左读都一样的数字。最大的由两个两位数乘积构成的回文数是 9009 = 91 * 99。
# 找出最大的有由个三位数乘积构成的回文数。
def huiwen(x):
y = str(x)[::-1]
if int(y) == x:
      return True
else:
      return False

max_Num = 0
for i in range(100, 999):
for j in range(100, 999):
      x = j * i
      if huiwen(x):
         if max_Num < x:
            max_Num = x

print(max_Num)

有马_冬巳 · 发表于 2020-10-2 12:16:19

'''一个回文数指的是从左向右和从右向左读都一样的数字。最大的由两个两位数乘积构成的回文数是 9009 = 91 * 99。
找出最大的由两个三位数的乘积构成的回文数。'''
palindrome = []
def ispalindrome(num):
s = str(num)
a = 0
b = len(s)-1
check = 0
for i in range(int((len(s)/2))):
if len(s) % 2 != 0:
return False
if s[a] != s[b]:
if check == 0:
check += 1
if check != 0:
break
a += 1
b -= 1
if check == 0:
return True
def multiply(num1,num2):
for number1 in range(num1, 99, -1):
for number2 in range(num2, 99, -1):
number = number1*number2
if ispalindrome(number):
palindrome.append([number1, number2, number])
def getbiggestpalindrome():
temp = 0
re = 0
for i in range(len(palindrome)-1):
if temp < palindrome[i][2]:
temp = palindrome[i][2]
re = i
print(palindrome[re])
start_palindrome = time.time()
multiply(999,999)
getbiggestpalindrome()
time_palindrome = time.time() - start_palindrome
print("%f秒" %time_palindrome)

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[993, 913, 906609]
0.712786秒

gonorth · 发表于 2020-11-4 22:29:43

def palin(x):
if str(x) == str(x)[::-1]:
      return True
return False

def getp():
list1 = [0]
list2 = [0,0]
for i in range(100,1000):
      for j in range (100,1000):
         if palin(i*j):
            x = i * j
            if x > list1[0]:
                  list1.pop()
                  list1.append(x)
                  if (i + j )> (list2[0] + list2[1]):
                     list2.pop()
                     list2.pop()
                     list2.append(i)
                     list2.append(j)

print (list2)
print (list1)
getp()

xg-sco · 发表于 2020-12-26 23:19:49

#include <stdio.h>
#define NUM 1000
int main ()
{
int max_1 = -1024, max_2 = 0, max_3 = 0;
int multiply, i, j;
int a, b, c, d, e, f;
for (i = 900; i < NUM; i++)
{
for (j = 900; j < i; j++)
{
multiply = i * j;
a = multiply / 100000 % 10;
b = multiply / 10000 % 10;
c = multiply / 1000 % 10;
d = multiply / 100 % 10;
e = multiply / 10 % 10;
f = multiply % 10;
if (a == f && b == e && c == d)
{
if (multiply > max_1)
{
max_1 = multiply;
max_2 = j;
max_3 = i;
}
// printf("%d = %d * %d\n", multiply, i, j);
}
}
}
printf("%d = %d * %d\n", max_1, max_2, max_3);
return 0;
}

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a1351468657 · 发表于 2021-3-6 13:23:21

#include <stdio.h>
#include <math.h>
main()
{
int i, j, k, num, mun = 0, temp;
for (i = 998001; i > 10000; i--)
{
num = i;
for (j = 5; j >= 0; j--)
{
temp = num % 10;
mun += (temp * pow(10, j));
num = (num - temp) / 10;
}
if (mun == i)
{
for (k = 999; k > 100; k--)
{
if (mun % k == 0)
{
temp = mun / k;
if ((temp / 1000) == 0)
{
goto Label;
}
}
}
}
mun = 0;
}
Label:printf("%d = %d * %d\n", mun, k, mun / k);
}

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永恒的蓝色梦想 · 发表于 2021-3-10 22:45:29

本帖最后由永恒的蓝色梦想于 2021-3-10 22:50 编辑

constexpr bool isPalindrome(unsigned int val)noexcept {
unsigned int copy = val, rev = 0;
while (val) {
rev += val;
val /= 10;
rev -= val * 10;
}
return !(copy ^ rev);
}

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hjg · 发表于 2021-4-1 21:11:54

无名侠发表于 2015-7-9 14:41
难点应该就是判断是否为回文了。
我用的方法有些歪门邪道，利用字符串来判断，不过效率还行。
然后就是三 ...

906609

hjg · 发表于 2021-4-2 08:56:15

906609

public static void main(String[] args) {
System.out.println(largestPalindrome(3));
}
public static int largestPalindrome(int n) {
if(n == 1) return 9;
//计算给定位数的最大值
long max = (long)Math.pow(10,n) - 1;
//从max - 1开始循环，原因见上文
for(long i = max - 1; i > max / 10; i--){
//1. 构造回文数
String s1 = String.valueOf(i);
long rev = Long.parseLong(s1 + new StringBuilder(s1).reverse().toString());
//2. 检验该回文数能否由给定的数相乘得到
for(long x = max; x * x >= rev; x --){
if(rev % x == 0) return (int)(rev );
}
}
return -1;
}

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酷爱语言的小白 · 发表于 2021-4-26 14:20:34

def number(x):
y = 10**x-1
z = 10**(x-1)
for i in range(y**2,z**2,-1):
j = int(str(i)[::-1])
if i == j:
for m in range(y,z,-1):
if i% m == 0:
n = int(i/ m)
if n in range(z,y):
return m, n, i
print(number(3))

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Kuri5u · 发表于 2021-5-3 20:51:40

#由于909*902=819918，我们从这里开始。
target=[]
for a in range(909,1000):
for b in range(902,1000):
r=a*b#r for result
#显然所求的数为六位数
if (str(r)[0]==str(r)[5])and (str(r)[1]==str(r)[4]) and (str(r)[2]==str(r)[3]):
target.append(r)
print(target)

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结果为：
[819918, 906609, 824428, 886688, 861168, 888888, 861168, 888888, 886688, 906609]

michaelwatson · 发表于 2021-5-6 16:53:12

本帖最后由 michaelwatson 于 2021-5-6 16:55 编辑

好啊

中国好青年 · 发表于 2021-5-18 15:42:52

a = []

for m in range(99,1000):
for n in range(99,1000):
      i = m*n
      b = str(i)
      c = list(b)
      if c[::-1] == c:
         a.append(i)

print(max(a))
答案906609

13425808254 · 发表于 2021-7-15 19:53:36

a = 0
b = ''
for i in range(100,1000):
   for o in range(100,1000):
         if str(i*o) == str(i*o)[::-1]:
               if i*o > a:
                     a = i*o
                     b = "(" + str(i) + ',' + str(o) + ')'
               else:
                     continue
         else:
               continue
else:
   print(a)
   print(b)

卢本伟牛逼 · 发表于 2021-8-8 22:52:51

#include <stdio.h>
int count;
int ispalindrome(int n)
{
int result = 0;
int value = n;
while(n!=0)
{
count++;
result = (n%10) + result * 10;
n = n / 10;
}
return (result == value)?1:0;
}
int main(void)
{
int max_i, max_j;
int max;
int i=999,j=999;
for(i=999;i>=100;i--)
{
for(j=999;j>=100;j--)
{
if (i<max_i && j < max_j) break; //不加这段count = 4718843加上count = 500591
if (ispalindrome(i * j)){
if (i * j > max){
max_i = i;
max_j = j;
max = (i * j > max)?i * j:max;
}
}
}
}
printf("%d * %d = %d\n", max_i, max_j, max);
printf("count = %d\n", count);
return 0;
}

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鱼塘里的鱼儿 · 发表于 2021-10-6 09:15:32

#include <iostream>
using namespace std;
int main()
{
int a,b,c;
for(int i=100;i<=999;i++)
{
a=i%10;
b=i/10%10;
c=i/100;
if(a==c)
cout<<i<<endl;
}
return 0;
}
//C++写的

ft215378 · 发表于 2021-10-8 16:18:07

#最大的两个三位数乘积构成的回文数

#检查是否回文
def check(n, start, end):
if start > end:
      return 1
else:
      return check(n, start+1, end-1) if n[start] == n[end] else 0

#提取数字计算
result = []
res = []
for a in range(100, 999):
for b in range(500,999):
      string = str(a * b)
      length = len(string) - 1
      if check(string, 0, length):
         result.append(int(string))
         if max(result) == int(string):
            res.append((a,b))

print(res[-1])

番杰 · 发表于 2021-10-27 13:42:09

本帖最后由番杰于 2021-10-27 13:58 编辑

#include<stdio.h>
int mian(void)
{
int t1,t10,t100,temp;
int a,b,max;
int flag = 0 ;
for(int i = 997799;i>10000;i--) // 从997799开始，是因为它是998001（999*999）之下最大的回文数
{
if(i>100000) //六位回文数
{
t1 = i % 10;
t10 = (i % 100) / 10;
t100 = (i % 1000) /100;
temp = (t1 * 100) + (t10 * 10) + t100;
if(temp == (i/1000)) //判断是否是回文数
{
for(int j = 999;j>=100;j--)
{
if(i % j == 0)
{
if(j > 99 && j < 1000)
{
a = j ;
b = i / j ;
max = i;
flag = 1;
break;
}
}
}
if(flag == 1)
break;
}
}
else //五位回文数
{
t1 = i % 10;
t10 = (i % 100) / 10;
temp = (t1 * 100) + (t10 * 10) ;
if(temp == (i/1000))
{
for(int j = 999;j>=100;j--)
{
if(i % j == 0)
{
if(j > 99 && j < 1000)
{
a = j ;
b = i / j ;
max = i;
flag = 1;
break;
}
}
}
if(flag == 1)
break;
}
}
printf("%d = %d + %d",c,a,b);
return 0 ;
}

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learner-ray · 发表于 2021-11-13 02:10:38

最大回文数为：993 * 913 = 906609

#include<stdio.h>
int main(void)
{
int i, j;
int mul, lum, max = 0;
int k, buf;
int a, b;
for (i = 999; i > 100; i--)
{
for (j = 999; j > 100; j--)
{
lum = 0;
mul = i * j;
buf = mul;
while (mul)
{
k = mul % 10;
mul = mul / 10;
lum = lum * 10 + k;
}
if (lum == buf)
{
if (lum > max)
{
max = lum;
a = i; b = j;
}
}
}
}
printf("最大回文数为：%d * %d = %d\n", a, b, max);
return 0;
}

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题目4：找出由两个三位数乘积构成的回文

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