The sum problem Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21416    Accepted Submission(s): 6287

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]
//题意很好理解，就不说了，在这块说一下思路吧：
1、首先得知道等差公式；
2、此题是公差为1的等差数列；
3、此题的n,m很大（不可能模拟所有的起点和终点）；
技巧转换：
只模拟起点和数列的长度，知道起点i和数列长度j，很容易根据等差数列公式求得这个子序列的和：(i+(i+j-1))*j/2;
化简公式得：
j*j+(2*i-1)=2*m;因为i>=1,m已知，所以j的取值为：j<=sqrt(2*m);
所以知道这个规律就可以写这个题了。 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<map>
#include<math.h>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 10010
#define M 1000000007
using namespace std;
int main()
{
	int n,m,i,j;
	while(scanf("%d%d",&n,&m),n|m)
	{
		for(j=sqrt(2*m);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;
}