判断一个数字的N!个，是否可以被奇数整除。
原题描述
https://codeforces.com/contest/2043/problem/B
Educational Codeforces Round 173 (Rated for Div. 2)
B.Digits
Artem wrote the digit d on the board exactly n! (for example:5!=1*2*3*4*5 ) times in a row.
So,he got the number dddd....ddd(exactly n! digits)
Now he is curious about which odd digits form 1 to 9 divide the number written on the board.
(odd:1,3,5,7,9 in(1 to 9 % 2!=0),
even:2,4,8,10(1 to 10) %2==0)
Input
The first line contains a single integer t(1<=t<=100)-the numbers of test cases.
The next t tests cases follow.
Each test case consists of a sigle line containing two integers n and d((2≤n≤109, 1≤d≤9).
Output
For each test case,output the odd digits in ascending order that divide the number written on the board.
Example
Input
3
2 6
7 1
8 5
OutputCopy
1 3
1 3 7 9
1 3 5 7 9
Note
The factorial of a positive integer n(n!) is the product of all integers from 1 to n.
For example, the factorial of 5 is 1*2*3*4*5=120
思路：
首先需要知道的知识
1.N阶层=N!=1*2*3*....N,例如5！ is 1*2*3*4*5=120
2 多个1，2，3.....9 整除的情况。
（1）3！=6个1，2，3....9可以整除3
（2）3！=6个1，2，3....9可以整除7
（3）3！=6个1，2，3....9 并且整除3，9的也可以
（4）6！的数字1，2，3...9 可以整除9
3 奇数odd=整除2有余数
根据以上规则来编写
STEP1 获得10以内的odd奇数，放在列表里list
STEP2 判断是否%1，3，5，7，9 每一种情况，一共5个情况，分别输出即可。
当1的时候，整除1==0即可
当3的时候，整除3==0 或者3！=6个单数，也是可以整除3，7的，这里就输出这个整除数
当5，仅仅1个
当7，类似3
当9，需要注意：整除9，6!也可以，还有3！并且%3 也是可以。
cpp
/*
https://codeforces.com/contest/2043/problem/B
Educational Codeforces Round 173 (Rated for Div. 2)
B.Digits
Artem wrote the digit d on the board exactly n! (for example:5!=1*2*3*4*5 ) times in a row.
So,he got the number dddd....ddd(exactly n! digits)
Now he is curious about which odd digits form 1 to 9 divide the number written on the board.
(odd:1,3,5,7,9 in(1 to 9 % 2!=0),
even:2,4,8,10(1 to 10) %2==0)
Input
The first line contains a single integer t(1<=t<=100)-the numbers of test cases.
The next t tests cases follow.
Each test case consists of a sigle line containing two integers n and d((2≤n≤109, 1≤d≤9).
Output
For each test case,output the odd digits in ascending order that divide the number written on the board.
Example
Input
3
2 6
7 1
8 5
OutputCopy
1 3
1 3 7 9
1 3 5 7 9
Note
The factorial of a positive integer n(n!) is the product of all integers from 1 to n.
For example, the factorial of 5 is 1*2*3*4*5=120
*/
#include <iostream>
#include <string>
using namespace std;
//ref others
void d_divide_odd()
{
int n = 0, d = 0;
cin >> n >> d;
char space1 = ' ';
int index = 0;
int *list = new int[5];
//get odd : 1,3,5,7,9
for (int i = 1; i <=9; i = i + 2)
{
list[index] = i;
index++;
}
//show
//1 in(1,2,3,4,5,6,7,8,9)
if (d % 1==0)
{
cout << list[0] << space1;
}
// case : 3
//1) for:1..9 has odd (3,6,9) %3 == 0
//2) new math knowledge: the factorial of n! = 3!=1*2*3= 6 times of d = %3 == 0
if (d%3 == 0 || n>=3)
{ //3!=1*2*3=6
cout << list[1] << space1;
}
// case : 5
//only one 5 in(1,2,3,4,5,6,7,8,9)
if (d % 5 == 0)
{
cout << list[2] << space1;
}
// case : 7
//1) only one 7 in(1,2,3,4,5,6,7,8,9)
//2) new math knowledge: the factorial of n! = 3!=1*2*3= 6 times of d = %7 == 0
if (d % 7 == 0 || n >= 3)
{
cout << list[3] << space1;
}
//Note: you can try to make 1,2,3...9 of n!(1,2,3)times % ==0
// case : 9
// 9 is 3,6,9
if (d % 9==0 || n >=6 || (d%3 == 0 && n >= 3) )
{
cout << list[4] << space1;
}
//Note: you can try to make 1,2,3...9 of n!(1,2,3,4,5,6)times % 9 ==0
cout << endl;
}
int main()
{
int t = 0;
cin >> t;
while (t--)
{
d_divide_odd();
}
return 0;
}完成，这里需要学习的新知识就是，3！%3，7==0，另外%3 并且N>=3！也是可以整除9