java - C语言的一道关于循环的练习题求解

Question

做这道题目没什么思路？不知道该怎么下手,求大神讲解一下

Answer 1

http://blog.csdn.net/houyp520...

Answer 2

The simplest thing is to exhaustively enumerate the arrangements of 9 numbers.

Optimization is to eliminate some impossible situations first. For example, the middle number is a multiple of 2, and the last number is a multiple of 3. And so on.

Answer 3

Just write recursion. . Or is it called iteration?

Answer 4

public class Sidney {

public static void main(String[] args) {
    int[] s = new int[9];
    for (int i = 300; i < 999; i+=3) {
        int flag = 1;
        Set<Integer> set = new HashSet<>();
        int a = i / 3;
        int b = (i / 3) * 2;
        s[0]=a%10;s[1]=a%100/10;s[2]=a/100;
        s[3]=b%10;s[4]=b%100/10;s[5]=b/100;
        s[6]=i%10;s[7]=i%100/10;s[8]=i/100;
        for (int i1 : s) {
            if (!set.add(i1) || i1 == 0) {
                flag = 0;
                break;
            }
        }
        if (flag == 1) {
            System.out.println(a + "\t" + b + "\t" + i);
        }
    }
}

}

Answer 5

Liu Rujia’s question.

Enumerate the first number, which is the number accounting for 1 in 1:2:3.

The enumeration range is 123~345. Based on this number, calculate the other two numbers, and then determine whether exactly 9 numbers are used.

Answer 6

The idea is very simple. Since you want the ratio to be 1:2:3 and each number needs to be used once, then directly enlarge 1, 2, and 3 x times, and then determine whether each number appears only once. 1 requires less than 100 times of magnification to reach 3 digits, so just start from 123. code show as below.
int testa()
{

int a=1,b=2,c =3;
char szA[10]={0},szB[4]={0},szC[4]={0},cTag[10]={0};

for(int i=123;i<=333;i++)
{
    memset(szA,0x00,10);memset(szB,0x00,4);memset(szC,0x00,4);memset(cTag,0x00,10);
    sprintf_s(szA,"%d",a*i);sprintf_s(szB,"%d",b*i);sprintf_s(szC,"%d",c*i);

    strcat_s(szA,szB);strcat_s(szA,szC);
    int j=0;
    for(j=0;j<9;j++)
    {
        if(cTag[szA[j]-'1']!=0)
            break;
        cTag[szA[j]-'1']=1;
    }

    if(j==9)
        printf("%s\n",szA);
}

return 0;

}