• UVA 160


     Factors and Factorials 

    The factorial of a number N (written N!) is defined as the product of all the integers from 1 to N. It is often defined recursively as follows:

    displaymath27

    displaymath28

    Factorials grow very rapidly--5! = 120, 10! = 3,628,800. One way of specifying such large numbers is by specifying the number of times each prime number occurs in it, thus 825 could be specified as (0 1 2 0 1) meaning no twos, 1 three, 2 fives, no sevens and 1 eleven.

    Write a program that will read in a number N ( tex2html_wrap_inline39 ) and write out its factorial in terms of the numbers of the primes it contains.

    Input

    Input will consist of a series of lines, each line containing a single integer N. The file will be terminated by a line consisting of a single 0.

    Output

    Output will consist of a series of blocks of lines, one block for each line of the input. Each block will start with the number N, right justified in a field of width 3, and the characters `!', space, and `='. This will be followed by a list of the number of times each prime number occurs in N!.

    These should be right justified in fields of width 3 and each line (except the last of a block, which may be shorter) should contain fifteen numbers. Any lines after the first should be indented. Follow the layout of the example shown below exactly.

    Sample input

    5
    53
    0

    Sample output

      5! =  3  1  1
     53! = 49 23 12  8  4  4  3  2  2  1  1  1  1  1  1
            1
    #include <cstdio>
    #include <cstring>
    using namespace std;
    bool isprime(int a)
    {
    	int i;
    	for (i = 2; i*i <= a;i++)
    	if (a%i == 0)
    		return false;
    	return true;
    }
    int prime[100], count[100];
    int main()
    {
    	int n;
    	int i,num;
    	for (i = 2, num = 0; i <= 100;i++)
    	if (isprime(i))
    	{
    		prime[num++] = i;
    	}
    
    	while (scanf("%d", &n) == 1 && n)
    	{
    		memset(count,0,sizeof(count));
    		int maxn = 0;
    		for (i = 2; i <= n; i++)
    		{
    			int m = i;
    			int j;
    			for (j = 0; j < num; j++)
    			{
    				while (m%prime[j] == 0)
    				{
    					m = m / prime[j];
    					count[j]++;
    					if (j>maxn)
    						maxn = j;
    				}
    			}
    		}
    		printf("%3d! =", n);
    		for (i = 0; i <= maxn; i++)
    		{
    			if (i == 15)
    				printf("
          "); 
    			printf("%3d", count[i]);
    		}
    			
    		printf("
    ");
    	}
    	return 0;
    }
    

      这道题应该特别注意输出格式。

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  • 原文地址:https://www.cnblogs.com/lakeone/p/4149887.html
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