算法 – 求和为n的连续正整数序列（C++）

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//  求和为n的连续正整数序列 - C++ - by Chimomo
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//  题目： 输入一个正整数n，输出全部和为n的连续正整数序列。比如：输入15，因为1+2+3+4+5=4+5+6=7+8=15，所以输出3个连续序列1-5、4-6和7-8。
////  Answer: Suppose n = i+(i+1)+...+(j-1)+j, then n = (i+j)(j-i+1)/2 = (j*j-i*i+i+j)/2 => j^2+j+(i-i^2-2n) = 0 => j = (sqrt(1-4(i-i^2-2n))-1)/2 => j = (sqrt(4i^2+8n-4i+1)-1)/2.//          We know 1 <= i < j <= n/2+1, so for each i in [1,n/2], do this arithmetic to check if there is a integer answer.////  Note: 二次函数 ax^2+bx+c=0 的求根公式为： x = (-b±sqrt(b^2-4ac)) / 2a。////****************************************************************************************************#include <iostream>#include <cassert>#include <stack>#include <math.h>using namespace std ;int FindConsecutiveSequence(int n){	int count = 0;	for (int i = 1; i <= n/2; i++)	{		double sqroot = sqrt(4*i*i + 8*n - 4*i + 1);		int floor = sqroot;		if(sqroot == floor)		{			cout << i << "-" << (sqroot - 1) / 2 << endl;			count++;		}	}	return count;}int main(){	int count = FindConsecutiveSequence(15);	cout << "Totally " << count << " sequences found." << endl;	return 0;}// Output:/*1-54-67-8Totally 3 sequences found.*/