Timus 1120. Sum of Sequential Numbers 数学题

时间：2014-04-27 21:38:05 收藏：0 阅读：544

There is no involute formulation concerning factitiously activity of SKB Kontur in this problem. Moreover, there is no formulation at all.

Input

There is the only number N, 1 ≤ N ≤ 109.

Output

Your program is to output two positive integers A and P separated with a space such that:

N = A + (A + 1) + … + (A + P ? 1).
You are to choose a pair with the maximal possible value of P.

Sample

input	output
14	2 4

这里使用的数学思维：分解两个因子的思想，利用程序特点循环求解

发现自己的数学水平还是不行，要继续修补。-- 这些数学思想其实早就学过的，不过使用的不多，故此容易忘记。

N = A + (A + 1) + … + (A + P ? 1) 求出公式：N=(A+A+p-1)*p/2;

2*N=P*(2A+P-1) 然后是分解因子得到：x=P; y=(2A+P-1)

2*N=x*y

然后循环x，即p，求得符合条件的A就结束循环。

自己推导下，就明白啦。

void SumofSequentialNumbers1120()
{
	long long S = 0;
	cin>>S;
	S <<= 1;
	for (int p = (int)sqrtl(S); p >= 0 ; p--)
	{
		if (S % p == 0)
		{
			int y = int(S / p);
			int a = y + 1 - p;
			if (a % 2 == 0)
			{
				cout<<a/2<<‘ ‘<<p;
				break;
			}
		}
	}
}

Timus 1120. Sum of Sequential Numbers 数学题,码迷,mamicode.com