n값은 찾아냈는데 m의 규칙성은 못 찾음....
결국 방정식으로 계산....
/*
Subject: #138 Street NUmbers
e-mail: morcavon@gmail.com
Homepage: http://www.morcavon.com
Building: 01/04/2005 ~ 02/04/2005
Last Update: 04/04/2005
*/
/*
<OUTPUT>
find 10 pairs of 1 + 2 + 3..+ n = n + (n+1) + (n+2)..+ m
*/
#include <iostream>
using namespace std;
#include <cmath>
double sigma(double i, double n);
inline void output(double n, double m);
bool search(const double m);
int main()
{
#if 1
double n, m, pre, t;
int cnt = 0;
cout.setf(ios::fixed);
cout.precision(0);
for (n = 1, m = 1, pre = 0; cnt < 10; cnt++) {
t = n * 6 - pre; // pre: 두 단계 전의 n값
pre = n;
n = t;
// m값의 규칙성은 찾기 힘들다=_= 윽..그냥 계산하는 수밖에...
// ∑1 to n == ∑n to m
output(n, (-1 + sqrt(1 + 8*n*n)) / 2);
}
#endif
return 0;
}
// brutal force =_=;;;;; 나오긴 나옴
bool search(const double m)
{
double left = 1, right = m, mid;
while (left <= right) {
mid = (int)((left + right) / 2);
if (sigma(1, mid) > sigma(mid, m)) {
right = mid - 1;
}
else if (sigma(1, mid) < sigma(mid, m)) {
left = mid + 1;
}
else
{ // equal
output(mid, m);
return true;
}
}
return false;
}
double sigma(double i, double n)
{
return ((i + n) * (n - i + 1)) / 2; // 등차수열(?)
}
inline void output(double n, double m)
{
cout.setf(ios::fixed);cout.precision(0);
cout.width(10);cout << n;
cout.width(10);cout << m << endl;
}
