Problem 4 | Project Euler | 欧

[sx349]

https://pe-cn.github.io/4/

Problem 4 Largest palindrome productA 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 m

[Archer0v0]

# 999 999
# 998 899
# ...
for i in range(999, 100, -1):
    num_temp = i%10*100 + (i//10)%10*10 + i//100
    num = i * 1000 + num_temp
    for i in range(999, 100, -1):
        if num%i == 0:
            if num//i <= 999 and num//i >= 100:
                print(i, num//i, num)
                break
    else:
        continue
    break

[killex2012]

def huiwen( N ):
    strN = str(N)
    lens = int(len(strN))
    half = int(lens / 2)
    for x in range(half ):
        if strN[x] != strN[-x-1]:
            return False
    return True
max = 0
for x in range(999,100,-1):
    for y in range(999,100,-1):
        z = x * y
        if huiwen(z):
            if max < z:
                print("{} * {} = {} 是回文".format(x, y, z))
                max = z
print(max)

---

995 * 583 = 580085 是回文
993 * 913 = 906609 是回文
906609

[summerrightnow]

public class Problem4 { public static void main(String[] args) { String s; for(int i = 998001;i >= 100000;i--) { s = String.valueOf(i); if(s.charAt(0) == s.charAt(5) && s.charAt(1) == s.charAt(4) && s.charAt(2) == s.charAt(3)) { // System.out.println(i); for(int j = 100;j <= 999;j++) { int k = i / j; int len = String.valueOf(k).length(); if(i % j == 0 && len == 3) { System.out.println(i + "," + j + "," + k); return; } } } } } }

[Stardusten]

Mathematica

先写个判断回文数的谓词：

palindromeQ[x_]:=(IntegerDigits@x === Reverse@IntegerDigits@x)

用 Outer 构建所有三位数相乘可能的结果构成的列表，可以用 DeleteDuplicates 去重：

Outer[Times, Range[100, 999], Range[100, 999]] //Flatten //DeleteDuplicates

再筛一遍，取最大即可

Cases[%, _?palindromeQ] //Max

用时 0.34s （用 AbsoluteTiming 计时）

Haskell

palindromeQ :: Int -> Bool
palindromeQ n = n_digits == reverse n_digits
    where n_digits = show n

main = do
    print . maximum $ filter palindromeQ [x*y | x <- [100..999], y <- [100..999]]

用时 0.07s ，Haskell yyds!（终端使用 time 计时）

[QlieQueen]

using namespace std;

int func(int num) { int x = num, y = 0; while (num) { y = 10 * y + num % 10; num /= 10; } if (x == y) return 1; return 0; }

int main() { int ans = 1; for (int i = 100; i <= 999; i++) { for (int j = 100; j <= 999; j++) { if (func(i j) && ans < (i j)) { ans = i * j; } } } cout << ans << endl; return 0; }

[QlieQueen]

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int func(int num) {
    int x = num, y = 0;
    while (num) {
        y = 10 * y + num % 10;
        num /= 10;
    }
    if (x == y) return 1;
    return 0;
}

int main() {
    int ans = 1;
    for (int i = 100; i <= 999; i++) {
        for (int j = 100; j <= 999; j++) {
            if (func(i * j) && ans < (i * j)) {
                ans = i * j;
            }
        }
    }
    cout << ans << endl;
    return 0;
}

[Yuan-yuan1]

include <bits/stdc++.h>

using namespace std; int a[7]; int recycle(int i) {int m; for( m=0;;m++) { if(i==0) break; if(i>0) {a[m]=i%10; i=i/10; }

} for(int q=0;q<=(m-1)/2;q++) { if(a[q]!=a[m-1-q]) return 0; }

return 1;

} int main() { int ans; for(int i=999;i>=100;i--) { for(int j=999;j>=i;j--) {if(recycle(ij)&&ans<ij) { ans=i*j; } }

} cout<<ans<<endl; return 0; }

[ujqt007]

def is_palindrome(n): if str(n) == str(n)[::-1]: return True

a, b, max = 100, 100, 10000 for i in range(100,1000): for j in range(100, 1000): p = i*j if is_palindrome(p): if p >= max: a, b, max =i, j, p

print(a,b,max)

[zhiduiniyouganjue]

include

using namespace std;

int func(int x) { int raw = x, t = 0; while(x) { t = t * 10 + x % 10; x /= 10; } return raw == t; }

int main() { int ans = 0; for (int i = 100; i < 1000; i++) { for (int j = i; j < 1000; j++) { if (func(i j)) { ans = max(ans, i j); } } } cout << ans << endl; return 0; }

[HCX0426]

include

using namespace std;

int func(int x) { int raw = x, t = 0; while (x) { t = t * 10 + x % 10; x /= 10; } return raw == t; }

int main() { int ans = 0; for (int i = 100; i < 1000; i++) { for (int j = i; j < 1000; j++) { if (func(i j)) { ans = max(ans, i j); } } } cout << ans << endl; return 0; }

[llll-bit]

include

bool huiwen(int n){

int a[6];          
int j=0;
int i=0;
while(n!=0){

     if(n/10!=0){

    a[i]=n%10;
    n=n/10;
    i++;
   }else{

    a[i]=n;
    n=n/10;
   }
}

while(j<=i){

if(a[j]==a[i-j]){

    j++;
}else{

return false;
}

} return true; }

int main(){

int res=0;

for(int i=100;i<1000;i++){

for(int j=100;j<1000;j++){

    if(huiwen(i*j)){

       if(i*j>=res){

        res=i*j;
       }
    }
}

}

printf("%d",res);
return 0;

}

[ghost]

python3

print(max((i for i in (x*y for x in range(100,999) for y in range(100,999)) if str(i)[::-1]==str(i))))

[Little-Chen-T]

include <bits/stdc++.h>

using namespace std;

stringstream ss;

string s;

bool isPalindrome (int x){

ss << x;
ss >> s;
ss.clear();

int len = s.size();

for (int i = 0; i < len/2; ++i){
    if (s[i] != s[len-i-1]) return false;
}

return true;

}

int main(){

vector<int> v;

for (int i = 999;i >= 100; --i){
    for (int j = 999; j >= 100; --j){
        if (isPalindrome(i*j)) v.push_back(i*j);
    }
}

sort(v.begin(), v.begin()+v.size());

cout << v[v.size()-1];

return 0;

}

[pastgonex]

// 最大回文乘积

#include <iostream>

using namespace std;

// ①
bool check(int x) {
    int t = x;
    int digit = 0;
    while (t) {
        digit = digit * 10 + t % 10;
        t /= 10;
    }
    return digit == x;
}

int main() {
    // int ans = 0;
    // for (int i = 100; i <= 999; i++) {
    //     for (int j = 100; j <= 999; j++) {
    //         int t = i * j;
    //         if (check(t)) {
    //             ans = max(ans, t);
    //         }
    //     }
    // }
    // cout << ans << endl;

    // `abccba` = 10^5a + 10^4b + 10^3c + 10^2c + 10b + a
    //          = 100000a + 10000b + 1000c + 100c + 10b + a
    //          = 100001a + 10010b + 1100c
    //          = 100001a + 10010b + 1100c
    //          = 11(9091a + 910b + 100c)
    // 因此这个6位数是11的倍数
    int ans = 0;
    for (int i = 999; i >= 100; i--) {
        for (int j = 999; j >= 100; j--) {
            int t = i * j;
            if (t % 11 ||!check(t)) continue;
            ans = max(t, ans);
        }
    }
    cout << ans << endl;
    return 0;
} 

[zhou-ying-wan]

include

include

using namespace std;

int a;

int equality(int x) { int p = x,t=0; while(x) { t = t*10+x%10; x/=10; }

return t == p;

}

int main() { int res = 0; for(int i = 100;i<1000;i++) { for(int j = i;j<1000;j++) { a=i*j; if(equality(a)) { res = max(res,a); } } }

cout<<res;

return 0;

}

[herenoflag]

for i in range(999,900,-1): for j in range(999,900,-1): n=i*j if str(n)==str(n)[::-1]: print(i,j,n) quit()

[ScandiumLin]

Mathematica

m = Table[a*b, {a, 100, 999}, {b, 100, 999}];

n = Table[ If[IntegerReverse[m[[i, j]]] == m[[i, j]], m[[i, j]], 0], {i, 1, 900}, {j, 1, 900}];

p = Table[Max[n[[k]]], {k, 1, 900}];

Max[p]

[JeromeGao0420]

include<bits/stdc++.h>

using namespace std; int m[10]; bool HuiWen(int num) { int count=0; int temp=num; while(temp!=0) { m[count++]=temp%10; temp/=10; } for(int i=0;i<count/2;++i) { if(m[i]!=m[count-1-i]) return false; } return true; } int main() { int result=0; int k; for(int i=101;i<=999;++i) { for(int j=101;j<=999;++j) { k=i*j;
if(HuiWen(k)) { if(k>result) result=k; } } } cout<<result; return 0; }

[hrizon123]

include

int func(int x) { int raw = x, t = 0; while (x != 0) { t = t * 10 + x % 10; x /= 10; } return raw == t; }

int main() { int ans = 0; for (int i = 100; i < 1000; i++) { for (int j = i; j < 1000; j++) { if (func(i j) && i j > ans) { ans = i * j; } } } printf("%d\n", ans); return 0; }

[p-gp]

#include <stdio.h>

_Bool isPa(int x);

int main()
{
    int x = 999;

    while (1) {
        if (isPa(x * x)) {
            printf("%d", x * x);
            break;
        }
        if (isPa(x * (x - 1))) {
                printf("%d", x * (x - 1));
                break;
        }
        --x;
    }

    return 0;
}

_Bool isPa(int x)
{
    int y = 0, z = x;
    do {
        y = y * 10 + x % 10;
    } while ((x /= 10) > 0);

    return z == y;
}

.....................................................................................................................................

698896

[wyj72660]

package oulajihua;

public class 回文数 { public static void main(String[] args) { int max = 0; for(int i = 100; i <= 999; i++) { for(int j = 100; j <= 999; j++) { if(max <= i j && isHuiWenShu(i j + "")) { max = i * j; } } } System.out.println(max); } public static boolean isHuiWenShu(String str) { StringBuffer sb = new StringBuffer(str); StringBuffer s2 = sb.reverse(); String s3 = s2.toString(); boolean flag = s3.equals(str); return flag; } }

[miaomiaopu]

// ans = 906609
class P4 {
    public static Integer flip(Integer i) {
        return Integer.parseInt(String.valueOf(new StringBuilder(String.valueOf(i)).reverse()));
    }

    public static void solve() {
        int max = 0;
        for (int i = 999; i >= 100; i--) {
            for (int j = 999; j >= 100; j--) {
                Integer tmp = i * j;
                Integer tmpFlip = flip(tmp);
                if (tmp.equals(tmpFlip)) {
                    max = Math.max(tmp, max);
                }
            }
        }
        System.out.println(max);
    }
}

[oldtommmy]

public class T04 {
    public static void main(String[] args) {
        int ans = 0;
        for (int i = 999; i >= 100; i--) {
            for (int j = 999; j >=100 ; j--) {
                if (isPal(i * j) && i*j > ans){
                    ans = i * j;
                }
            }
        }
        System.out.println(ans); //906609
    }
    public static boolean isPal(int n){
        StringBuffer s1 = new StringBuffer(String.valueOf(n));
        if (s1.reverse().toString().equals(String.valueOf(n))){
            return true;
        }
        return false;
    }
}

[2994276187]

include

include

using namespace std; int sum,res; int raw(int x) { int store = x, t = 0; while (1) { t = t 10 + x % 10; x /= 10; if (x == 0) break; } return t == store; } int main() { for (int i = 900;i <= 999;i++) { for (int j = 900;j <= 999;j++) { sum = i j; if (raw(sum)) res = i * j; } } cout << res; return 0; }

[ghost]

def palindrome_num (num): str_num = str(num) a = len(str_num) i = 0 while i <=a//2: if str_num[i] == str_num[a-1-i]: i += 1 else: return False break return True

p_num = [] for m in range(999,99,-1): for n in range(m,99,-1): num = m*n if palindrome_num(num): p_num.append(num) break print(max(p_num))

[182663823]

include

using namespace std;

int func(int x) { int t = 0, row = x; while(x) { t = t * 10 + x % 10; x /= 10; } return t == row; }

int main() { int ans = 0; for(int i = 100; i < 1000; i++) { for(int j = i ; j < 1000; j++) { int t = i * j; if(func(t)){ cout << t << endl; ans = max(ans,t); } } } cout << ans << endl; return 0; }

[Ouchan233]

Mathematica

Select[Flatten[Transpose[{Range[100, 999]}] . {Range[100, 999]}],
IntegerDigits[#] == Reverse[IntegerDigits[#]] &] // Max

Answer: 906609

[weisus]

Select[Table[i*j, {i, 100, 999}, {j, 100, 999}] // Flatten,
  IntegerReverse[#] == # &] // Max

[SaintGuilliman]

include<bits/stdc++.h>

define ll long long

bool check(ll x) { ll y=0,p=x; while(p) { y=y10+p%10; p=p/10; } return x==y; } int main() { ll ans=0; for(ll i=100;i<=999;i++) { for(ll j=100;j<=999;j++) { if(ij>ans&&check(ij))ans=ij; } } printf("%lld\n",ans); }

[LawrenceLee985]

偷懒，不想用数学方法

include<bits/stdc++.h>

using namespace std;

string reverse1(string a){ int left=0; int right=a.length()-1; while(left<right){ char temp=a[left]; a[left]=a[right]; a[right]=temp; left++; right--; } return a; }

int main(){ vectorv1; for(int i=100;i<1000;i++){ for(int j=100;j<=i;j++){ int tmp=i*j; string string1=to_string(tmp); if(string1==reverse1(string1)){ v1.push_back(tmp); } } } sort(v1.begin(),v1.end()); for(int i=0;i<v1.size();i++){ cout<<v1[i]<<endl; } return 0; }

[pear404]

include

include

int main() { int i,j,m,n=0,p; char a[7]; for(i=998001;i>0;i--) { m=i; itoa(i,a,10); if(a[0]==a[5]&&a[1]==a[4]&&a[2]==a[3]) { for(j=999;j>0;j--) { if(m%j==0) { if(j>100&&m/j<1000) { n=1; p=j; break; } else break; } } if(n==1) { printf("%d %d %s\n",j,m/j,a); break; } } } }

[Kevinzhou03]

include

include

bool isHuiwenOrNot(int *arr, int bit) { bool flag = true;

for (int i = 0; i < bit; i++)
{
    if (arr[i] != arr[bit - 1 - i])
        flag = false;
}

return flag;

}

int main() { int a, b; int maxSum = 0; for (int i = 100; i < 1000; i++) { for (int j = 100; j < 1000; j++) { int sum = i j; int bit = 0; int arr[6]; // the maximum bit is less than 6（999x999） while (sum != 0) { arr[bit] = sum % 10; sum /= 10; bit++; } if (isHuiwenOrNot(arr, bit) && ij > maxSum) { maxSum = i*j; a = i; b = j; }
} } printf("the maximum HuiWen number is %d, which is multiplied by %d and %d", maxSum, a, b); return 0; }

[adjname]

include

int main() { printf("%d",999*999); }

[adjname]

include

int main() { int max=111; for(int i=1;i<10;i++) for(int j=1;j<10;j++) for(int k=1;k<10;k++) { if(i==j&&j==k) { if(i(1010)+j(10)+k1>=max) max=i(1010)+j(10)+k1; } } printf("%d",max*max); }

[adjname]

include

int main() { int max=111; for(int i=1;i<10;i++) for(int j=1;j<10;j++) for(int k=1;k<10;k++) { if(i==j&&j==k) { if(i(1010)+j(10)+k1>=max) max=i(1010)+j(10)+k1; } } printf("%d",maxmax); }

[AdwardAllan]

Julia

global Sets = Set([])
for i ∈ 9:-1:1
    for j ∈ 9:-1:1
        for k ∈ 9:-1:1
            for l ∈ 9:-1:1
                pro = string((900+i*10+j)*(900+k*10+l))
                if pro == pro[end:-1:1]
                    global Sets = union(Sets,parse(Int,pro))
                end
            end
        end
    end
end

println("maximum palindrome product is ",maximum(Sets))

# maximum palindrome product is 906609

[Liaco123]

from count_time import count_time

def is_palidromem(num):
    return str(num) == str(num)[::-1]

@count_time
def largest_palindrome_product(num: int) -> int:
    largest_palindrome = 0
    num_range = range(10**num-1, 10**(num-1)-1, -1)
    for i in num_range:
        for j in num_range:
            res = i * j
            if res < largest_palindrome:
                break
            if is_palidromem(res) and res > largest_palindrome:
                largest_palindrome = res
                largest_i = i
                largest_j = j
    return f"{largest_palindrome} = {largest_i} x {largest_j}"

if __name__ == '__main__':
    largest_palindrome = largest_palindrome_product(3)
    print(largest_palindrome)

[OrangeGlaze]

#include <stdio.h>

#define MAX_N 1000

int isPalindromic(int num) {
    int _num = num, p = 0;
    while (_num) {
        p = p * 10 + _num % 10;
        _num /= 10;
    }
    return p == num;
}

int main() {
    int maxN = 0;
    for (int i = 100; i < MAX_N; i++) {
        for (int j = i; j < MAX_N; j++) {
            if (isPalindromic(i * j) && (i * j) > maxN) {
                maxN = i * j;
            }
        }
    }
    printf("%d\n", maxN);
    return 0;
}

[loveqiaoshen]

include

using namespace std;

/ Largest palindrome product 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. 最大回文乘积 回文数是指从前往后读和从后往前读都一样的数。由两个2位数相乘得到的回文数中，最大的是 9009 = 91 x 99 求最大的由两个3位数相乘得到的回文数 /

bool isHuiwen(int num) { int ori_num = num; int Hui_num = 0;

while (num > 0)
{
    Hui_num = Hui_num * 10 + num % 10;
    num /= 10;
}

return (Hui_num == ori_num);

}

int main() { int max = 0; for (int i = 999; i >= 100; i--) { for (int j = 999; j >= 100; j--) { if (isHuiwen(i j) && i j > max) { max = i * j; //break; } } }

cout << max << endl;

return 0;

}

[Culaccino00]

• 思路点拨：

  主要思考要点是，如何判断这个数是回文数，第一反应是转成字符串，然后首尾比对，看了其他人题解后发现可以用数学方法判断：

  原理大致为：任何一个数除以10的余数就是该数最后一位；任何一个数除以10的商就是排除掉最后一位后的数；所以 ，一个数1234 就可以通过这种方法得到 123 和 4；接下来对 123 进行同样的操作，就得到 12 和 3；接下来得到 1 和 2；接着得到 0 和 1；整个过程是个循环，当商不是 0 的时候就一直如此；每次拿到一个余数，都用来构造新数，新数=新数*10+余数；所以经过四次循环后，我们得到新数 4321 ，如果是回文，那么新数应该等于原数，否则，说明不是回文。

• 我的代码：

  #include <bits/stdc++.h>
  using namespace std;
  bool isPalindrome(int num)
  {
  int neww = 0;
  int old = num;
  while(num > 0)
  {
      neww = neww * 10 + num % 10;
      num /= 10;
  }
  return neww == old;
  }
  int main()
  {
  int maxx = 0;
  int num;
  for(int i = 100; i < 1000; i ++)
  {
      for(int j = i; j < 1000; j ++)
      {
          num = i * j;
          if(isPalindrome(num) && num > maxx)
          {
              maxx = num;
          }
      }
  }
  printf("%d", maxx);
  return 0;
  } 

[wkkkgk]

number_digits <- function(n) {

• 将数字转换为字符向量

• digits <- as.character(strsplit(as.character(n), "")[[1]])

• 转换回数值类型

• digits <- as.numeric(digits)
• if(digits==rev(digits))
• {
• return(1)
• }
• else
• return(0)
• }

  MAX<-0;#最大回文数 for(i in 100:999)

• {
• for(j in 100:999)
• {
• temp<-i*j
• if(number_digits(temp)&&temp>MAX)
• {
• MAX<-temp
• }
• }
• }