The Erdös-Straus Conjecture 题解

题面

Description

The Brocard Erdös-Straus conjecture is that for any integern > 2 , there are positive integersa ≤ b ≤ c,so that :

\[{4\over n}={1\over a}+{1\over b} +{1\over c} \]

There may be multiple solutions. For example:

\[{4\over 18}={1\over 9}+{1\over 10}+{1\over 90}={1\over 5}+{1\over 90}+{1\over 90}={1\over 5}+{1\over 46}+{1\over 2470} \]

Input

The first line of input contains a single decimal integer P, (1 ≤ p ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of a single line of input. It contains the data set number,K, followed by a single space, followed by the decimal integer n,(2 ≤ n ≤ 50000)

Output

For each data set there is one line of output. The single output line consists of the data set number, K, followed by a single space followed by the decimal integer values a, b and c in that order, separated by single spaces

Sample Input

5
1 13
2 529
3 49849
4 49850
5 18

Sample Output

1 4 18 468
2 133 23460 71764140
3 12463 207089366 11696183113896622
4 12463 310640276 96497380762715900
5 5 46 2070

题解

对于这个式子

\[\frac{4}{n}=\frac{1}a+\frac1b+\frac1c \]

不如先解除\(a\)的范围，首先\(a\)必然大于\(\frac{n}4\),因为\(b和c\)项不能为0，其次\(a\)要小于等于\(\frac{3*n}{4}\)，因为要满足字典序最小，\(a\)必然最小。然后我们在这个范围枚举\(a\)，计算可行的\(b和c\)，不如将\(b和c\)看成一个整体，这样可以解出\(\frac{1}{b}+\frac{1}{c}\)

\[\frac{1}b+\frac{1}c=\frac{4*i-n}{n*i}(\frac{n}{4} +1\leq i \leq \frac{3*n}{4}) \]

如果解出后化简得到的分数正好分子为1，那么我们就可以直接得到\(b和c\)，我们假设解出的分数为\(\frac{x}{y},x=1\),那么b和c就可以拆分为

\[b=\frac{1}{y+1} \\ c=\frac{1}{(y+1)*y} \]

如果化简后不为1，则从小到大枚举b，找到最小的c，枚举下界从解出的分数的倒数+1开始，保证\(\frac{1}{b} < \frac{x}{y}\)到倒数的两倍结束，因为如果超过两倍的倒数b就大于c了，这时应轮换b和c使字典序最小，找出答案记录立刻退出循环即可

代码写的比较丑

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {
	return a % b == 0 ? b : gcd(b, a % b);
}
int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		int k; ll n;
		scanf("%d%lld", &k, &n);
		ll l = n / 4 + 1;
		ll r = n * 3 / 4;
		ll a, b, c;
		bool flag = false;
		for (ll i = l; i <= r; i++) {
			if (flag) break;
			ll x = n * i;
			ll y = 4 * i - n;
			ll num = gcd(x, y);
			x /= num;
			y /= num;
			if (y == 1) {
				a = i;
				b = x + 1;
				c = (x + 1) * x;
				flag = true;
				break;
			}
			else {
				ll s = x / y + 1;
				ll t = 2 * x / y;
				for (ll j = s; j <= t; j++) {
					ll tmp = y * j - x;
					if ((x * j) % tmp == 0) {
						a = i;
						b = j;
						c = x * j / tmp;
						flag = true;
						break;
					}
				}
			}
		}
		printf("%d %lld %lld %lld\n", k, a, b, c);
	}
	return 0;
}

最近可能都没太有空更题解了，没做的题有点多，作业还好多wwww