Description
Mr. Port plans to start a new business renting one or more floors of the new skyscraper with one giga floors, MinatoHarukas. He wants to rent as many vertically adjacent floors as possible, because he wants to show advertisement on as many vertically adjacent windows as possible. The rent for one floor is proportional to the floor number, that is, the rent per month for the n-th floor is n times that of the first floor. Here, the ground floor is called the first floor in the American style, and basement floors are out of consideration for the renting. In order to help Mr. Port, you should write a program that computes the vertically adjacent floors satisfying his requirement and whose total rental cost per month is *exactly equal* to his budget.For example, when his budget is 15 units, with one unit being the rent of the first floor, there are four possible rent plans, 1+2+3+4+5, 4+5+6, 7+8, and 15. For all of them, the sums are equal to 15. Of course in this example the rent of maximal number of the floors is that of 1+2+3+4+5, that is, the rent from the first floor to the fifth floor.
Input
The input consists of multiple datasets, each in the following format.
- b
A dataset consists of one line, the budget of Mr. Port b as multiples of the rent of the first floor. b is a positive integer satisfying 1 < b < (10^9). The end of the input is indicated by a line containing a zero. The number of datasets does not exceed 1000.
Output
For each dataset, output a single line containing two positive integers representing the plan with the maximal number of vertically adjacent floors with its rent price exactly equal to the budget of Mr. Port. The first should be the lowest floor number and the second should be the number of floors.
样例输入
15
16
2
3
9699690
223092870
847288609
900660121
987698769
999999999
0
样例输出
1 5
16 1
2 1
1 2
16 4389
129 20995
4112949 206
15006 30011
46887 17718
163837 5994
大意就是找自然数列里和为n的最长的一段。设这一段起始的数为a，长度为l，那么和为(a imes l +frac{l imes (l-1)}{2})，注意到l大致在(sqrt{n})范围内（把l换为(sqrt{n})的话离着n已经很接近了），不妨扩大一下（比如(3 imes sqrt{n})）然后从大到小枚举i，找到一组符合的直接break即可。
记得开long long！
#include <bits/stdc++.h>
#define int long long
using namespace std;
int n;
signed main()
{
while(scanf("%lld", &n) && n)
{
for(int i = sqrt(n) * 3; i >=1; i--)
{
int a = (n - i*(i - 1) / 2) / i;
if(a * i + i * (i - 1) / 2 == n && a > 0)
{
cout << a << ' ' << i << endl;
break;
}
}
}
}