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////  求和為n的連續正整數序列 - C++ - by Chimomo////  題目： 輸入一個正整數n，輸出所有和為n的連續正整數序列。例如：輸入15，由於1+2+3+4+5=4+5+6=7+8=15，所以輸出3個連續序列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: 二次函式 ax^2+bx+c=0 的求根公式為： 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.*/

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