请用C++解决以下问题:It is easy to see that for every fraction in the form 1/k(k > 0)(形如 1/k 的分数), we can always find two positive integers(正整数) x and y, x ≥ y, such that:1/k= 1/x+1/y.Now our question is: can you write a program(程序) that counts(计算) how many such pairs of x and y there are for any given k? Input: Input contains(包含) no more than 100 lines, each giving a value of k (0 < k ≤ 10000). Output: For each k, output the number(数量) of corresponding (x, y) pairs((x, y)关系对), followed by a sorted list(排列的列表) 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
时间: 2023-03-31 20:03:59 浏览: 128
我可以用C语言解决这个问题。以下是代码:
#include <stdio.h>
int main() {
int k;
while (scanf("%d", &k) == 1) {
int count = ;
printf("%d ", count);
for (int y = k + 1; y <= 2 * k; y++) {
int x = (k * y) / (y - k);
if (k * (x + y) == x * y) {
printf("1/%d = 1/%d + 1/%d ", k, x, y);
count++;
}
}
printf("\n");
}
return ;
}
对于每个给定的k,程序会输出对应的(x, y)关系对的数量,以及按顺序排列的x和y的值。程序的思路是,对于每个y,计算出对应的x,然后判断是否满足1/k=1/x+1/y的关系。如果满足,就输出对应的(x, y)关系对,并将计数器加1。最后输出计数器的值即可。