The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8287    Accepted Submission(s): 2530


Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.
Input Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

Output For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

Sample Input 20 10 50 30 0 0
Sample Output [1,4] [10,10] [4,8] [6,9] [9,11] [30,30]

我本來的想法是考慮子列的起點和終點，分別以s和e表示，由等差數列求和公式有（s+e）*（e-s+1）/2==M（1式），化為e*(e+1)-s*(s-1)==2*M, so , e=(int)sqrt(2*M+s*(s-1)),將得到的e再代回1式，成立則[s,e]滿足條件。

但是，2*M+s*(s-1)太大……

後來參考網上的一個演算法，不考慮子列的終點，而是考慮子列的起點和子列元素的個數，分別記為i，j。由等差數列求和公式，得(i+(i+j-1))*j/2==M ，即（2*i+j-1）*j/2==M（2式），故得i=（2*M/j-j+1）/2，將i，j代回2式，成立則[i,i+j-1]滿足條件。注意j最小為1，而由2式，得（j+2*i）*j=2*M，而i>=1，故j*j<=(int)sqrt(2*M).

AC code:

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	int N,M,i,j;
	while(scanf("%d%d",&N,&M) && M+N)
	{ 
	 	for(j=pow(2.0*M,0.5);j>0;j--)
		{
			i=(2*M/j-j+1)/2;
			if(j*(j+2*i-1)/2==M) printf("[%d,%d]\n",i,i+j-1);
		}
		printf("\n");					  
 	} 	
	return 0;	
}