Fractions Again?!

最新推荐文章于 2021-06-14 19:05:50 发布

原创最新推荐文章于 2021-06-14 19:05:50 发布 · 777 阅读

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#简单枚举

Fractions Again?!

It is easy to see that for every fraction in the form 1/k (k > 0), we can always find two positive integers x and y, x >= y, such that:

.
Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?

Input

Input contains no more than 100 lines, each giving a value of k (0 < k <= 10000).

Output

For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.

Sample Input

2
12

Sample Output

2
1/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24

需要枚举x、y，但是它们在一个什么范围内呢？我们注意到，1/x<=1/y，则1/k-1/y<=1/y，即y<=2*k 我们只需枚举y，再求出x即可。但是需要注意的是求x时，我开始的时候用的是：x=（double）1/(1/k-1/y)，结果溢出，因为括号里面仍然是整数的运算显然不对，但是改成x=1/((double)1/K-(double)1/y)之后虽不溢出，结果是错的，因为这里大量运用了除法，精度出了问题。改成x=（double）K*y/(y-K）就通过了，所以计算的时候应该避免进行大量的除法运算。
代码如下：

#include<cstdio>
#include<cmath>
using namespace std;

const int maxn=10000+5;
int Y[maxn],X[maxn];

int main()
{
	int K;
	while(scanf("%d",&K)==1)
	{
		int count=0;
		for(int y=K+1;y<=2*K;y++)
		{
			double x;
			x=(double)K*y/(y-K);

			if(x==floor(x))
			{
				++count;
				X[count]=(int)x;
				Y[count]=y;
			} 
		}
		printf("%d\n",count);
		for(int i=1;i<=count;i++)
			printf("1/%d = 1/%d + 1/%d\n",K,X[i],Y[i]);
	}
	return 0;
}