题面
Description
The Brocard Erdös-Straus conjecture is that for any integern > 2 , there are positive integersa ≤ b ≤ c,so that :
There may be multiple solutions. For example:
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
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})
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
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
", 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