Find all factorial numbers less than or equal to N

标签：printf   task   call   first   turn   log   clu   输出   应该   

A number N is called a factorial number if it is the factorial of a positive integer. For example, the first few factorial numbers are 1, 2, 6, 24, 120, …
Given a number N, the task is to print all factorial numbers smaller than or equal to N.

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains a number N as input.

Output:
For each test case, print all factorial numbers smaller than or equal to N in new line.

Constraints:
1<=T<=100
1<=N<=10¹⁸

Example:
Input:
2
2
6

Output:
1 2
1 2 6

下面是我的代码实现：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int num,i;
    scanf("%d",&num);
    for(i=0;i<num;i++)
    {
        int N,j,temp=1;
        scanf("%d",&N);
        for(j=1;j<=N;j++)
        {
            temp=temp*j;
            if(temp<=N)
                printf("%d ",temp);
        }
    }
    return 0;
}

　　这个程序现在还是错误的。因为我不确定输入的数值大小，只知道范围是：1<=T<=1001<=N<=10¹⁸，然后如果是最大的输入，这个输出是相当的大，那么这个应该是动态的，但是C里面怎么具体实现在输入都不确定的情况下，输出还是不确定的，我还没找到对应的关系。

如果换成C++语言，那直接就可以只用函数实现，C好难啊！

Find all factorial numbers less than or equal to N

标签：printf   task   call   first   turn   log   clu   输出   应该   

原文地址：http://www.cnblogs.com/wongyi/p/7680074.html