@EraYes 你渴望点赞吗？

def find_combinations(): results = [] for a in range(100, 334): b = 2 * a c = 3 * a combined = str(a) + str(b) + str(c) if len(combined) == 9 and set(combined) == set('123456789'): results.append(f"{a} {b} {c}") return results

combinations = find_combinations() for combo in combinations: print(combo)

kc2班yyy小恐龙高达1000分谁敢挑战

C++和Python都可以使用内置全排列来实现本题。

C++中algorithm库提供了next_permutation()用来生成排列

C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    vector<int> digits = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    while (next_permutation(digits.begin(), digits.end())){
        int num1 = digits[0] * 100 + digits[1] * 10 + digits[2];
        int num2 = digits[3] * 100 + digits[4] * 10 + digits[5];
        int num3 = digits[6] * 100 + digits[7] * 10 + digits[8];
        if (num2 == 2 * num1 && num3 == 3 * num1)
            printf("%d    %d    %d\n", num1, num2, num3);
    }
    return 0;
}

Python中itertools库提供了全排列迭代器permutations

from typing import List
from itertools import permutations

digits: List[int] = list(range(1, 10))
for perm in permutations(digits):
    num1: int = perm[0] * 100 + perm[1] * 10 + perm[2];
    num2: int = perm[3] * 100 + perm[4] * 10 + perm[5];
    num3: int = perm[6] * 100 + perm[7] * 10 + perm[8];
    if num2 == 2 * num1 and num3 == 3 * num1:
        print(f"{num1}    {num2}    {num3}")

当然，这里的全排列会有许多冗余的排列生成，比如num1 > 333 时，往后产生的数都不可能是答案，所以剩下的优化就交给你了。 😸

我是天才

#include

using namespace std;

int main() { cout<<"192 384 576"<<endl; cout<<"219 438 657"<<endl; cout<<"273 546 819"<<endl; cout<<"327 654 981"<<endl; return 0; }

a=[0]*10
for i in range(123,334):
    s=0
    for j in range(1,10):
        a[j]=0
    for k in range(1,4):
        n=k*i
        a[n%10]=1
        a[n//10%10]=1
        a[n//100]=1
    for m in range(1,10):
        s+=a[m]
    if s==9:
        print(str(i)+" "+str(2*i)+" "+str(3*i))

more easy

print("192 384 576\n219 438 657\n273 546 819\n327 654 981")